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how to access a address element from below nested json object IN PHP
Deleting an element from an array in PHPAdd new attribute (element) to JSON object using JavaScriptHow to get the client IP address in PHPReturning JSON from a PHP ScriptHow can I parse a JSON file with PHP?How do I turn a C# object into a JSON string in .NET?How do I POST JSON data with Curl from a terminal/commandline to Test Spring REST?Parse JSON code from Bing Map API in C#How do I pass variables and data from PHP to JavaScript?How to deserialize bing map json using C#?
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-2
I want access to addressLine, adminDistrict from address element which is stored in json format IN PHP
HK20271557
HK20271557
php json
asked Mar 28 at 12:52
Code warCode war
94 bronze badges
add a comment
|
-2
I want access to addressLine, adminDistrict from address element which is stored in json format IN PHP
HK20271557
HK20271557
php json
asked Mar 28 at 12:52
Code warCode war
94 bronze badges
assign that JSON to a variable in PHP. And use json_decode() on that variable
– Vipin Kumar Soni
Mar 28 at 13:16
why to use json_decode(),when it is already coming in json format
– Code war
Mar 28 at 13:23
to convert it in array
– Vipin Kumar Soni
Mar 28 at 13:24
$JsonData = json_decode($result,true), and how to access the addressline from the array
– Code war
Mar 28 at 13:25
check the answer...
– Vipin Kumar Soni
Mar 28 at 13:27
add a comment
|
-2
-2
-2
1
I want access to addressLine, adminDistrict from address element which is stored in json format IN PHP
HK20271557
HK20271557
php json
asked Mar 28 at 12:52
Code warCode war
94 bronze badges
I want access to addressLine, adminDistrict from address element which is stored in json format IN PHP
HK20271557
HK20271557
php json
php json
asked Mar 28 at 12:52
Code warCode war
94 bronze badges
asked Mar 28 at 12:52
Code warCode war
94 bronze badges
edited Mar 28 at 13:08
Code war
asked Mar 28 at 12:52
Code warCode war
94 bronze badges
asked Mar 28 at 12:52
Code warCode war
94 bronze badges
asked Mar 28 at 12:52
Code warCode war
94 bronze badges
94 bronze badges
assign that JSON to a variable in PHP. And use json_decode() on that variable
– Vipin Kumar Soni
Mar 28 at 13:16
why to use json_decode(),when it is already coming in json format
– Code war
Mar 28 at 13:23
to convert it in array
– Vipin Kumar Soni
Mar 28 at 13:24
$JsonData = json_decode($result,true), and how to access the addressline from the array
– Code war
Mar 28 at 13:25
check the answer...
– Vipin Kumar Soni
Mar 28 at 13:27
add a comment
|
assign that JSON to a variable in PHP. And use json_decode() on that variable
– Vipin Kumar Soni
Mar 28 at 13:16
why to use json_decode(),when it is already coming in json format
– Code war
Mar 28 at 13:23
to convert it in array
– Vipin Kumar Soni
Mar 28 at 13:24
$JsonData = json_decode($result,true), and how to access the addressline from the array
– Code war
Mar 28 at 13:25
check the answer...
– Vipin Kumar Soni
Mar 28 at 13:27
assign that JSON to a variable in PHP. And use json_decode() on that variable
– Vipin Kumar Soni
Mar 28 at 13:16
assign that JSON to a variable in PHP. And use json_decode() on that variable
– Vipin Kumar Soni
Mar 28 at 13:16
why to use json_decode(),when it is already coming in json format
– Code war
Mar 28 at 13:23
why to use json_decode(),when it is already coming in json format
– Code war
Mar 28 at 13:23
to convert it in array
– Vipin Kumar Soni
Mar 28 at 13:24
to convert it in array
– Vipin Kumar Soni
Mar 28 at 13:24
$JsonData = json_decode($result,true), and how to access the addressline from the array
– Code war
Mar 28 at 13:25
$JsonData = json_decode($result,true), and how to access the addressline from the array
– Code war
Mar 28 at 13:25
check the answer...
– Vipin Kumar Soni
Mar 28 at 13:27
check the answer...
– Vipin Kumar Soni
Mar 28 at 13:27
add a comment
|
2 Answers
2
active
oldest
votes
0
you need to access it by assigning it to the variable e.g. say
let a = HK01EAP000001D0"
then access it like this
console.log(a.resourceSets[0].resources[0].address)
answered Mar 28 at 13:02
Taj KhanTaj Khan
3804 silver badges16 bronze badges
They asked for it IN PHP
– Derek Pollard
Mar 28 at 13:10
add a comment
|
0
<?php
/*
if Array of JSON Object is returned then
Use following
*/
$json = '[HK20271557,
HK20271557]';
$array = json_decode($json);
foreach($array as $sub_array)
echo $sub_array->resourceSets[0]->resources[0]->address->addressLine . "<br/>";
echo $sub_array->resourceSets[0]->resources[0]->address->adminDistrict . "<br/>";
/*
And if single JSON Object is returned then
Use following
*/
$json = 'HK01EAP000001D0"
';
$JsonData = json_decode($json);
echo $JsonData->resourceSets[0]->resources[0]->address->addressLine . "<br/>";
echo $JsonData->resourceSets[0]->resources[0]->address->adminDistrict . "<br/>";
answered Mar 28 at 13:27
Vipin Kumar SoniVipin Kumar Soni
7418 silver badges18 bronze badges
$JsonData = json_decode($result,true); foreach ($JsonData as $sub_array) echo $sub_array->resourceSets[0]->resources[0]->address->addressLine . "<br/>"; echo $sub_array->resourceSets[0]->resources[0]->address->adminDistrict . "<br/>"; blank page is coming
– Code war
Mar 28 at 13:36
Please check that I have made the json response as array.Notice The Square brackets and comma in between json objects.
– Vipin Kumar Soni
Mar 28 at 13:39
but that is coming from an api, how can i change it,into the above format
– Code war
Mar 28 at 13:41
Is the API is returning in that exact format. And please remove the second parmeter from json_decode($result,true) and change it to json_decode($result).
– Vipin Kumar Soni
Mar 28 at 13:42
Is the API is returning one json Object OR the whole that you have given in Question
– Vipin Kumar Soni
Mar 28 at 13:45
|
show 3 more comments
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
you need to access it by assigning it to the variable e.g. say
let a = HK01EAP000001D0"
then access it like this
console.log(a.resourceSets[0].resources[0].address)
answered Mar 28 at 13:02
Taj KhanTaj Khan
3804 silver badges16 bronze badges
They asked for it IN PHP
– Derek Pollard
Mar 28 at 13:10
add a comment
|
0
you need to access it by assigning it to the variable e.g. say
let a = HK01EAP000001D0"
then access it like this
console.log(a.resourceSets[0].resources[0].address)
answered Mar 28 at 13:02
Taj KhanTaj Khan
3804 silver badges16 bronze badges
They asked for it IN PHP
– Derek Pollard
Mar 28 at 13:10
add a comment
|
0
0
0
you need to access it by assigning it to the variable e.g. say
let a = HK01EAP000001D0"
then access it like this
console.log(a.resourceSets[0].resources[0].address)
answered Mar 28 at 13:02
Taj KhanTaj Khan
3804 silver badges16 bronze badges
you need to access it by assigning it to the variable e.g. say
let a = HK01EAP000001D0"
then access it like this
console.log(a.resourceSets[0].resources[0].address)
answered Mar 28 at 13:02
Taj KhanTaj Khan
3804 silver badges16 bronze badges
answered Mar 28 at 13:02
Taj KhanTaj Khan
3804 silver badges16 bronze badges
answered Mar 28 at 13:02
Taj KhanTaj Khan
3804 silver badges16 bronze badges
answered Mar 28 at 13:02
Taj KhanTaj Khan
3804 silver badges16 bronze badges
3804 silver badges16 bronze badges
They asked for it IN PHP
– Derek Pollard
Mar 28 at 13:10
add a comment
|
They asked for it IN PHP
– Derek Pollard
Mar 28 at 13:10
They asked for it IN PHP
– Derek Pollard
Mar 28 at 13:10
They asked for it IN PHP
– Derek Pollard
Mar 28 at 13:10
add a comment
|
0
<?php
/*
if Array of JSON Object is returned then
Use following
*/
$json = '[HK20271557,
HK20271557]';
$array = json_decode($json);
foreach($array as $sub_array)
echo $sub_array->resourceSets[0]->resources[0]->address->addressLine . "<br/>";
echo $sub_array->resourceSets[0]->resources[0]->address->adminDistrict . "<br/>";
/*
And if single JSON Object is returned then
Use following
*/
$json = 'HK01EAP000001D0"
';
$JsonData = json_decode($json);
echo $JsonData->resourceSets[0]->resources[0]->address->addressLine . "<br/>";
echo $JsonData->resourceSets[0]->resources[0]->address->adminDistrict . "<br/>";
answered Mar 28 at 13:27
Vipin Kumar SoniVipin Kumar Soni
7418 silver badges18 bronze badges
$JsonData = json_decode($result,true); foreach ($JsonData as $sub_array) echo $sub_array->resourceSets[0]->resources[0]->address->addressLine . "<br/>"; echo $sub_array->resourceSets[0]->resources[0]->address->adminDistrict . "<br/>"; blank page is coming
– Code war
Mar 28 at 13:36
Please check that I have made the json response as array.Notice The Square brackets and comma in between json objects.
– Vipin Kumar Soni
Mar 28 at 13:39
but that is coming from an api, how can i change it,into the above format
– Code war
Mar 28 at 13:41
Is the API is returning in that exact format. And please remove the second parmeter from json_decode($result,true) and change it to json_decode($result).
– Vipin Kumar Soni
Mar 28 at 13:42
Is the API is returning one json Object OR the whole that you have given in Question
– Vipin Kumar Soni
Mar 28 at 13:45
|
show 3 more comments
0
<?php
/*
if Array of JSON Object is returned then
Use following
*/
$json = '[HK20271557,
HK20271557]';
$array = json_decode($json);
foreach($array as $sub_array)
echo $sub_array->resourceSets[0]->resources[0]->address->addressLine . "<br/>";
echo $sub_array->resourceSets[0]->resources[0]->address->adminDistrict . "<br/>";
/*
And if single JSON Object is returned then
Use following
*/
$json = 'HK01EAP000001D0"
';
$JsonData = json_decode($json);
echo $JsonData->resourceSets[0]->resources[0]->address->addressLine . "<br/>";
echo $JsonData->resourceSets[0]->resources[0]->address->adminDistrict . "<br/>";
answered Mar 28 at 13:27
Vipin Kumar SoniVipin Kumar Soni
7418 silver badges18 bronze badges
$JsonData = json_decode($result,true); foreach ($JsonData as $sub_array) echo $sub_array->resourceSets[0]->resources[0]->address->addressLine . "<br/>"; echo $sub_array->resourceSets[0]->resources[0]->address->adminDistrict . "<br/>"; blank page is coming
– Code war
Mar 28 at 13:36
Please check that I have made the json response as array.Notice The Square brackets and comma in between json objects.
– Vipin Kumar Soni
Mar 28 at 13:39
but that is coming from an api, how can i change it,into the above format
– Code war
Mar 28 at 13:41
Is the API is returning in that exact format. And please remove the second parmeter from json_decode($result,true) and change it to json_decode($result).
– Vipin Kumar Soni
Mar 28 at 13:42
Is the API is returning one json Object OR the whole that you have given in Question
– Vipin Kumar Soni
Mar 28 at 13:45
|
show 3 more comments
0
0
0
<?php
/*
if Array of JSON Object is returned then
Use following
*/
$json = '[HK20271557,
HK20271557]';
$array = json_decode($json);
foreach($array as $sub_array)
echo $sub_array->resourceSets[0]->resources[0]->address->addressLine . "<br/>";
echo $sub_array->resourceSets[0]->resources[0]->address->adminDistrict . "<br/>";
/*
And if single JSON Object is returned then
Use following
*/
$json = 'HK01EAP000001D0"
';
$JsonData = json_decode($json);
echo $JsonData->resourceSets[0]->resources[0]->address->addressLine . "<br/>";
echo $JsonData->resourceSets[0]->resources[0]->address->adminDistrict . "<br/>";
answered Mar 28 at 13:27
Vipin Kumar SoniVipin Kumar Soni
7418 silver badges18 bronze badges
<?php
/*
if Array of JSON Object is returned then
Use following
*/
$json = '[HK20271557,
HK20271557]';
$array = json_decode($json);
foreach($array as $sub_array)
echo $sub_array->resourceSets[0]->resources[0]->address->addressLine . "<br/>";
echo $sub_array->resourceSets[0]->resources[0]->address->adminDistrict . "<br/>";
/*
And if single JSON Object is returned then
Use following
*/
$json = 'HK01EAP000001D0"
';
$JsonData = json_decode($json);
echo $JsonData->resourceSets[0]->resources[0]->address->addressLine . "<br/>";
echo $JsonData->resourceSets[0]->resources[0]->address->adminDistrict . "<br/>";
answered Mar 28 at 13:27
Vipin Kumar SoniVipin Kumar Soni
7418 silver badges18 bronze badges
edited Mar 28 at 13:54
answered Mar 28 at 13:27
Vipin Kumar SoniVipin Kumar Soni
7418 silver badges18 bronze badges
answered Mar 28 at 13:27
Vipin Kumar SoniVipin Kumar Soni
7418 silver badges18 bronze badges
answered Mar 28 at 13:27
Vipin Kumar SoniVipin Kumar Soni
7418 silver badges18 bronze badges
7418 silver badges18 bronze badges
$JsonData = json_decode($result,true); foreach ($JsonData as $sub_array) echo $sub_array->resourceSets[0]->resources[0]->address->addressLine . "<br/>"; echo $sub_array->resourceSets[0]->resources[0]->address->adminDistrict . "<br/>"; blank page is coming
– Code war
Mar 28 at 13:36
Please check that I have made the json response as array.Notice The Square brackets and comma in between json objects.
– Vipin Kumar Soni
Mar 28 at 13:39
but that is coming from an api, how can i change it,into the above format
– Code war
Mar 28 at 13:41
Is the API is returning in that exact format. And please remove the second parmeter from json_decode($result,true) and change it to json_decode($result).
– Vipin Kumar Soni
Mar 28 at 13:42
Is the API is returning one json Object OR the whole that you have given in Question
– Vipin Kumar Soni
Mar 28 at 13:45
|
show 3 more comments
$JsonData = json_decode($result,true); foreach ($JsonData as $sub_array) echo $sub_array->resourceSets[0]->resources[0]->address->addressLine . "<br/>"; echo $sub_array->resourceSets[0]->resources[0]->address->adminDistrict . "<br/>"; blank page is coming
– Code war
Mar 28 at 13:36
Please check that I have made the json response as array.Notice The Square brackets and comma in between json objects.
– Vipin Kumar Soni
Mar 28 at 13:39
but that is coming from an api, how can i change it,into the above format
– Code war
Mar 28 at 13:41
Is the API is returning in that exact format. And please remove the second parmeter from json_decode($result,true) and change it to json_decode($result).
– Vipin Kumar Soni
Mar 28 at 13:42
Is the API is returning one json Object OR the whole that you have given in Question
– Vipin Kumar Soni
Mar 28 at 13:45
$JsonData = json_decode($result,true); foreach ($JsonData as $sub_array) echo $sub_array->resourceSets[0]->resources[0]->address->addressLine . "<br/>"; echo $sub_array->resourceSets[0]->resources[0]->address->adminDistrict . "<br/>"; blank page is coming
– Code war
Mar 28 at 13:36
$JsonData = json_decode($result,true); foreach ($JsonData as $sub_array) echo $sub_array->resourceSets[0]->resources[0]->address->addressLine . "<br/>"; echo $sub_array->resourceSets[0]->resources[0]->address->adminDistrict . "<br/>"; blank page is coming
– Code war
Mar 28 at 13:36
Please check that I have made the json response as array.Notice The Square brackets and comma in between json objects.
– Vipin Kumar Soni
Mar 28 at 13:39
Please check that I have made the json response as array.Notice The Square brackets and comma in between json objects.
– Vipin Kumar Soni
Mar 28 at 13:39
but that is coming from an api, how can i change it,into the above format
– Code war
Mar 28 at 13:41
but that is coming from an api, how can i change it,into the above format
– Code war
Mar 28 at 13:41
Is the API is returning in that exact format. And please remove the second parmeter from json_decode($result,true) and change it to json_decode($result).
– Vipin Kumar Soni
Mar 28 at 13:42
Is the API is returning in that exact format. And please remove the second parmeter from json_decode($result,true) and change it to json_decode($result).
– Vipin Kumar Soni
Mar 28 at 13:42
Is the API is returning one json Object OR the whole that you have given in Question
– Vipin Kumar Soni
Mar 28 at 13:45
Is the API is returning one json Object OR the whole that you have given in Question
– Vipin Kumar Soni
Mar 28 at 13:45
|
show 3 more comments
draft saved
draft discarded
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assign that JSON to a variable in PHP. And use json_decode() on that variable
– Vipin Kumar Soni
Mar 28 at 13:16
why to use json_decode(),when it is already coming in json format
– Code war
Mar 28 at 13:23
to convert it in array
– Vipin Kumar Soni
Mar 28 at 13:24
$JsonData = json_decode($result,true), and how to access the addressline from the array
– Code war
Mar 28 at 13:25
check the answer...
– Vipin Kumar Soni
Mar 28 at 13:27