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pass variables from template to views in django

Does Django scale?Using % url ??? % in django templatesDjango Generic Template URL ReferenceDjango urls.py not rendering correct templateDjango reverse lookup rendering the same (wrong) templatewrong (same) django template returned for every urlMezzanine ignores the view, displays the template but does not pass the contextDjango Got an unexpected keyword argument 'id_speaker'How to pass a variable from one view to another in DjangoDjango - template not being detected

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

0

On clicking on each location I have to fetch a text file from there. I need to pass these locations to the views.py to render the files.

template:

<script>
if(data.columns[j] == "outputFilePath")
 var op = JSON.stringify(data.tableData[i][j]);
 op = op.substring(1, op.length-1)
 row.append("<td><a href='/dhl/outputDir'>" + op + "</a></td>")

</script>

views. py:

def outputDir(request,location):
 text_data = open("location/stdout", "rb").read()
 return HttpResponse(text_data, content_type="text/plain")

urls.py

url(r'^dhl/outputDir',views.outputDir),

django django-views django-urls

share|improve this question

edited Mar 27 at 10:48

Johan

2,62811 gold badge77 silver badges1919 bronze badges

asked Mar 27 at 10:16

Shilpa NarayananShilpa Narayanan

311 bronze badge

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Are you sure that's from your template file? Because that looks more like a mix of php and python, and not proper jinja2-syntax
  
  – Johan
  Mar 27 at 10:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  yes that is from a html file, code snippet here is the javascript part of the html
  
  – Shilpa Narayanan
  Mar 27 at 10:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I see, I took the liberty to update the template code so it's a bit more obvious what happens :) However, can you emphasize what you need help with in your question? Currently your question states what you want to achieve, but it's not clear what you need help with. Take a minute to read How do I ask a good question? if you need help for how to formulate your question and code.
  
  – Johan
  Mar 27 at 10:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  How do I pass the variable 'op' to outputDir(views.py) so that I can fetch a file from the location---> text_data=open("op/stdout","rb").read() is possible
  
  – Shilpa Narayanan
  Mar 27 at 11:39
  

  
  
  
  

add a comment | 

0

On clicking on each location I have to fetch a text file from there. I need to pass these locations to the views.py to render the files.

template:

<script>
if(data.columns[j] == "outputFilePath")
 var op = JSON.stringify(data.tableData[i][j]);
 op = op.substring(1, op.length-1)
 row.append("<td><a href='/dhl/outputDir'>" + op + "</a></td>")

</script>

views. py:

def outputDir(request,location):
 text_data = open("location/stdout", "rb").read()
 return HttpResponse(text_data, content_type="text/plain")

urls.py

url(r'^dhl/outputDir',views.outputDir),

django django-views django-urls

share|improve this question

edited Mar 27 at 10:48

Johan

2,62811 gold badge77 silver badges1919 bronze badges

asked Mar 27 at 10:16

Shilpa NarayananShilpa Narayanan

311 bronze badge

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Are you sure that's from your template file? Because that looks more like a mix of php and python, and not proper jinja2-syntax
  
  – Johan
  Mar 27 at 10:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  yes that is from a html file, code snippet here is the javascript part of the html
  
  – Shilpa Narayanan
  Mar 27 at 10:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I see, I took the liberty to update the template code so it's a bit more obvious what happens :) However, can you emphasize what you need help with in your question? Currently your question states what you want to achieve, but it's not clear what you need help with. Take a minute to read How do I ask a good question? if you need help for how to formulate your question and code.
  
  – Johan
  Mar 27 at 10:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  How do I pass the variable 'op' to outputDir(views.py) so that I can fetch a file from the location---> text_data=open("op/stdout","rb").read() is possible
  
  – Shilpa Narayanan
  Mar 27 at 11:39
  

  
  
  
  

add a comment | 

On clicking on each location I have to fetch a text file from there. I need to pass these locations to the views.py to render the files.

template:

<script>
if(data.columns[j] == "outputFilePath")
 var op = JSON.stringify(data.tableData[i][j]);
 op = op.substring(1, op.length-1)
 row.append("<td><a href='/dhl/outputDir'>" + op + "</a></td>")

</script>

views. py:

def outputDir(request,location):
 text_data = open("location/stdout", "rb").read()
 return HttpResponse(text_data, content_type="text/plain")

urls.py

url(r'^dhl/outputDir',views.outputDir),

django django-views django-urls

share|improve this question

edited Mar 27 at 10:48

Johan

2,62811 gold badge77 silver badges1919 bronze badges

asked Mar 27 at 10:16

Shilpa NarayananShilpa Narayanan

311 bronze badge

<script>
if(data.columns[j] == "outputFilePath")
 var op = JSON.stringify(data.tableData[i][j]);
 op = op.substring(1, op.length-1)
 row.append("<td><a href='/dhl/outputDir'>" + op + "</a></td>")

</script>

def outputDir(request,location):
 text_data = open("location/stdout", "rb").read()
 return HttpResponse(text_data, content_type="text/plain")

url(r'^dhl/outputDir',views.outputDir),

share|improve this question

edited Mar 27 at 10:48

Johan

2,62811 gold badge77 silver badges1919 bronze badges

asked Mar 27 at 10:16

Shilpa NarayananShilpa Narayanan

311 bronze badge

share|improve this question

edited Mar 27 at 10:48

Johan

2,62811 gold badge77 silver badges1919 bronze badges

asked Mar 27 at 10:16

Shilpa NarayananShilpa Narayanan

311 bronze badge

edited Mar 27 at 10:48

Johan

2,62811 gold badge77 silver badges1919 bronze badges

edited Mar 27 at 10:48

Johan

2,62811 gold badge77 silver badges1919 bronze badges

asked Mar 27 at 10:16

Shilpa NarayananShilpa Narayanan

311 bronze badge

asked Mar 27 at 10:16

Shilpa NarayananShilpa Narayanan

311 bronze badge

1 Answer
1

active

oldest

votes

0

You can use basic template tags and default view functionality by passing parameters.

However, what you're trying to achieve, will make the view wide open for anyone to access folders in your application by entering any folder they want. You could add a list of allowed locations, which I have done in my solution below.

Template

if(data.columns[j] == "outputFilePath")
 var op = JSON.stringify(data.tableData[i][j]);
 op = op.substring(1, op.length-1)
 row.append("<td><a href='/dhl/outputDir/" + op + "'>" + op + "</a></td>")

views.py

def outputDir(request, location):
 # Make sure to check if the location is in the list of allowed locations
 if location in allowed_locations:
 text_data = open(location + "/stdout", "rb").read()
 return HttpResponse(text_data, content_type="text/plain")
 else:
 return PermissionDenied

You also need to add a parameter to the url:

urls.py

url(r'^dhl/outputDir/(?P<location>w+)', views.outputDir),

share|improve this answer

answered Mar 27 at 12:05

JohanJohan

2,62811 gold badge77 silver badges1919 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks Johan, the above method worked perfectly
  
  – Shilpa Narayanan
  Apr 1 at 9:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @ShilpaNarayanan No problem! If you feel it was an adequate answer that helped you, consider marking it as an accepted solution. Or if you want to be open for more possible answers you can just leave it as it is :)
  
  – Johan
  Apr 1 at 10:16
  

  
  
  
  

add a comment | 

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0

You can use basic template tags and default view functionality by passing parameters.

However, what you're trying to achieve, will make the view wide open for anyone to access folders in your application by entering any folder they want. You could add a list of allowed locations, which I have done in my solution below.

Template

if(data.columns[j] == "outputFilePath")
 var op = JSON.stringify(data.tableData[i][j]);
 op = op.substring(1, op.length-1)
 row.append("<td><a href='/dhl/outputDir/" + op + "'>" + op + "</a></td>")

views.py

def outputDir(request, location):
 # Make sure to check if the location is in the list of allowed locations
 if location in allowed_locations:
 text_data = open(location + "/stdout", "rb").read()
 return HttpResponse(text_data, content_type="text/plain")
 else:
 return PermissionDenied

You also need to add a parameter to the url:

urls.py

url(r'^dhl/outputDir/(?P<location>w+)', views.outputDir),

share|improve this answer

answered Mar 27 at 12:05

JohanJohan

2,62811 gold badge77 silver badges1919 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks Johan, the above method worked perfectly
  
  – Shilpa Narayanan
  Apr 1 at 9:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @ShilpaNarayanan No problem! If you feel it was an adequate answer that helped you, consider marking it as an accepted solution. Or if you want to be open for more possible answers you can just leave it as it is :)
  
  – Johan
  Apr 1 at 10:16
  

  
  
  
  

add a comment | 

0

You can use basic template tags and default view functionality by passing parameters.

However, what you're trying to achieve, will make the view wide open for anyone to access folders in your application by entering any folder they want. You could add a list of allowed locations, which I have done in my solution below.

Template

if(data.columns[j] == "outputFilePath")
 var op = JSON.stringify(data.tableData[i][j]);
 op = op.substring(1, op.length-1)
 row.append("<td><a href='/dhl/outputDir/" + op + "'>" + op + "</a></td>")

views.py

def outputDir(request, location):
 # Make sure to check if the location is in the list of allowed locations
 if location in allowed_locations:
 text_data = open(location + "/stdout", "rb").read()
 return HttpResponse(text_data, content_type="text/plain")
 else:
 return PermissionDenied

You also need to add a parameter to the url:

urls.py

url(r'^dhl/outputDir/(?P<location>w+)', views.outputDir),

share|improve this answer

answered Mar 27 at 12:05

JohanJohan

2,62811 gold badge77 silver badges1919 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks Johan, the above method worked perfectly
  
  – Shilpa Narayanan
  Apr 1 at 9:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @ShilpaNarayanan No problem! If you feel it was an adequate answer that helped you, consider marking it as an accepted solution. Or if you want to be open for more possible answers you can just leave it as it is :)
  
  – Johan
  Apr 1 at 10:16
  

  
  
  
  

add a comment | 

You can use basic template tags and default view functionality by passing parameters.

However, what you're trying to achieve, will make the view wide open for anyone to access folders in your application by entering any folder they want. You could add a list of allowed locations, which I have done in my solution below.

Template

if(data.columns[j] == "outputFilePath")
 var op = JSON.stringify(data.tableData[i][j]);
 op = op.substring(1, op.length-1)
 row.append("<td><a href='/dhl/outputDir/" + op + "'>" + op + "</a></td>")

views.py

def outputDir(request, location):
 # Make sure to check if the location is in the list of allowed locations
 if location in allowed_locations:
 text_data = open(location + "/stdout", "rb").read()
 return HttpResponse(text_data, content_type="text/plain")
 else:
 return PermissionDenied

You also need to add a parameter to the url:

urls.py

url(r'^dhl/outputDir/(?P<location>w+)', views.outputDir),

share|improve this answer

answered Mar 27 at 12:05

JohanJohan

2,62811 gold badge77 silver badges1919 bronze badges

if(data.columns[j] == "outputFilePath")
 var op = JSON.stringify(data.tableData[i][j]);
 op = op.substring(1, op.length-1)
 row.append("<td><a href='/dhl/outputDir/" + op + "'>" + op + "</a></td>")

def outputDir(request, location):
 # Make sure to check if the location is in the list of allowed locations
 if location in allowed_locations:
 text_data = open(location + "/stdout", "rb").read()
 return HttpResponse(text_data, content_type="text/plain")
 else:
 return PermissionDenied

url(r'^dhl/outputDir/(?P<location>w+)', views.outputDir),

share|improve this answer

answered Mar 27 at 12:05

JohanJohan

2,62811 gold badge77 silver badges1919 bronze badges

answered Mar 27 at 12:05

JohanJohan

2,62811 gold badge77 silver badges1919 bronze badges

answered Mar 27 at 12:05

JohanJohan

2,62811 gold badge77 silver badges1919 bronze badges