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load buttons via ajax and then after clicking loaded buttons,load further contents for the relevant button

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0

I want to load some content in form of buttons via ajax and then after clicking those loaded buttons I want to be able to load further contents for each button relevantly.

is this possible ?

Here is what I have till now !

Calling Function by onclick.

<button onclick="findWeek(<?php echo $row['week']; ?>)"></button>

This will be the div which will have the loaded content:

 <div id="stats">

 </div>

Function

 function findWeek(ide) 
 

 var xhr = new XMLHttpRequest();



 xhr.open("GET" ,"start.php?ide="+ide,true);

 xhr.send(null);


 xhr.onreadystatechange = function()
 if( (xhr.readyState == '4') && ( xhr.status == '200') )
 

 document.getElementById("stats").innerHTML = xhr.responseText;
 


 ;

 

The loaded content is from the file start.php.

Here is the code of start.php, this will be loaded and then will be feed to the div with id #stats.

When these buttons are loaded I want to be able to click on these buttons and load further contents that is related with each button.

start.php

if($run = mysqli_query($connect,"SELECT `day` from `foodd_schedule` WHERE `week` = '$week' group by day")){
 while($row = mysqli_fetch_assoc($run)){

 ?>

 <button class="btn btn-danger" onclick="findday()"><?php echo $row['day']; ?></button>

findday Function inside start.php file.

<script type="text/javascript">

function findday()
 alert("I'm called");

The findday() function is not even called and the console shows the error of findday() is not defined.

How to solve this problem.
Thanks.

javascript php ajax

share|improve this question

asked Mar 24 at 17:20

NprogrammerNprogrammer

84

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  start.php how do you get ide value? Also, for complex operations, try to test each function before adding more?
  
  – Marios Nikolaou
  Mar 24 at 21:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I've send ide when I called the fucntion findweek(), you see !, obviously I've that , that's why I've send them as an argument.
  
  – Nprogrammer
  Mar 25 at 5:36
  

  
  
  
  

add a comment | 

0

I want to load some content in form of buttons via ajax and then after clicking those loaded buttons I want to be able to load further contents for each button relevantly.

is this possible ?

Here is what I have till now !

Calling Function by onclick.

<button onclick="findWeek(<?php echo $row['week']; ?>)"></button>

This will be the div which will have the loaded content:

 <div id="stats">

 </div>

Function

 function findWeek(ide) 
 

 var xhr = new XMLHttpRequest();



 xhr.open("GET" ,"start.php?ide="+ide,true);

 xhr.send(null);


 xhr.onreadystatechange = function()
 if( (xhr.readyState == '4') && ( xhr.status == '200') )
 

 document.getElementById("stats").innerHTML = xhr.responseText;
 


 ;

 

The loaded content is from the file start.php.

Here is the code of start.php, this will be loaded and then will be feed to the div with id #stats.

When these buttons are loaded I want to be able to click on these buttons and load further contents that is related with each button.

start.php

if($run = mysqli_query($connect,"SELECT `day` from `foodd_schedule` WHERE `week` = '$week' group by day")){
 while($row = mysqli_fetch_assoc($run)){

 ?>

 <button class="btn btn-danger" onclick="findday()"><?php echo $row['day']; ?></button>

findday Function inside start.php file.

<script type="text/javascript">

function findday()
 alert("I'm called");

The findday() function is not even called and the console shows the error of findday() is not defined.

How to solve this problem.
Thanks.

javascript php ajax

share|improve this question

asked Mar 24 at 17:20

NprogrammerNprogrammer

84

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  start.php how do you get ide value? Also, for complex operations, try to test each function before adding more?
  
  – Marios Nikolaou
  Mar 24 at 21:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I've send ide when I called the fucntion findweek(), you see !, obviously I've that , that's why I've send them as an argument.
  
  – Nprogrammer
  Mar 25 at 5:36
  

  
  
  
  

add a comment | 

I want to load some content in form of buttons via ajax and then after clicking those loaded buttons I want to be able to load further contents for each button relevantly.

is this possible ?

Here is what I have till now !

Calling Function by onclick.

<button onclick="findWeek(<?php echo $row['week']; ?>)"></button>

This will be the div which will have the loaded content:

 <div id="stats">

 </div>

Function

 function findWeek(ide) 
 

 var xhr = new XMLHttpRequest();



 xhr.open("GET" ,"start.php?ide="+ide,true);

 xhr.send(null);


 xhr.onreadystatechange = function()
 if( (xhr.readyState == '4') && ( xhr.status == '200') )
 

 document.getElementById("stats").innerHTML = xhr.responseText;
 


 ;

 

The loaded content is from the file start.php.

Here is the code of start.php, this will be loaded and then will be feed to the div with id #stats.

When these buttons are loaded I want to be able to click on these buttons and load further contents that is related with each button.

start.php

if($run = mysqli_query($connect,"SELECT `day` from `foodd_schedule` WHERE `week` = '$week' group by day")){
 while($row = mysqli_fetch_assoc($run)){

 ?>

 <button class="btn btn-danger" onclick="findday()"><?php echo $row['day']; ?></button>

findday Function inside start.php file.

<script type="text/javascript">

function findday()
 alert("I'm called");

The findday() function is not even called and the console shows the error of findday() is not defined.

How to solve this problem.
Thanks.

javascript php ajax

share|improve this question

asked Mar 24 at 17:20

NprogrammerNprogrammer

84

<button onclick="findWeek(<?php echo $row['week']; ?>)"></button>

 <div id="stats">

 </div>

 function findWeek(ide) 
 

 var xhr = new XMLHttpRequest();



 xhr.open("GET" ,"start.php?ide="+ide,true);

 xhr.send(null);


 xhr.onreadystatechange = function()
 if( (xhr.readyState == '4') && ( xhr.status == '200') )
 

 document.getElementById("stats").innerHTML = xhr.responseText;
 


 ;

 

if($run = mysqli_query($connect,"SELECT `day` from `foodd_schedule` WHERE `week` = '$week' group by day")){
 while($row = mysqli_fetch_assoc($run)){

 ?>

 <button class="btn btn-danger" onclick="findday()"><?php echo $row['day']; ?></button>

<script type="text/javascript">

function findday()
 alert("I'm called");

share|improve this question

asked Mar 24 at 17:20

NprogrammerNprogrammer

84

share|improve this question

asked Mar 24 at 17:20

NprogrammerNprogrammer

84

asked Mar 24 at 17:20

NprogrammerNprogrammer

84

asked Mar 24 at 17:20

NprogrammerNprogrammer

84