Hello experts, I am new in django and trying to learn how to build django web-framework for MySQL database.I can post my query (search term) and get desired results. But I am trying to modify my project so user can submit query in submission page and see their query parameter in URL when it is executed. Something like this: submission page: http://localhost:8000/ and after execution page will be like this:http://localhost:8000/xtrack/?searchid=XXXX
But still now I couldn't figure out how to do it in a right way after spending few days.
forms.py
from django import forms
from models import Query
class SQLForm(forms.ModelForm):
xtrackid=forms.CharField(max_length=100)
def checkxID(self):
xtrackid=self.cleaned_data.get("xtrackid")
return xtrackid
class QueryForm(forms.ModelForm):
class Meta:
model=Query
fields=["xtrackid"]
views.py
from django.shortcuts import render
from django.http import HttpResponse
from forms import SQLForm, QueryForm
import sys
def search_form(request):
return render(request, 'index.html')
def search(request):
form = QueryForm(request.POST or None)
if form.is_valid():
instance = form.save(commit=False)
xtrackid = form.cleaned_data.get("xtrackid")
xtrackid =xtrackid.strip()
conn = MySQLdb.connect (host = "localhost", user = "root", passwd = "XXXX", db = "XXXtracker")
cursor = conn.cursor ()
cursor.execute ("SELECT xInfo.xtideID, xIDunID.AccessionNumber FROM xInfo, xIDunID WHERE xInfo.xtideID = xIDunID.xtideID AND xIDunID.xtideID LIKE '%" + xtrackid +"%'")
row = cursor.fetchone ()
listrow= list(row)
contextres={}
if cursor.rowcount==0:
contexterror = {
'outputerror': xtrackid
}
return render(request, 'errorform.html', contexterror)
else:
if contextres.has_key(str(listrow[0])):
contextres[str(listrow[0])].append(listrow[1])
else:
contextres[str(listrow[0])]= [listrow[1]]
resulstdict = {'contextresultinfo': contextres}
return render(request, 'resultform.html', {'xinfo': resulstdict, 'query': xtrackid})
conn.close()
else:
return HttpResponse('Please submit a valid search term.')
urls.py
from django.conf.urls import include, url
from django.contrib import admin
from myapp import views
urlpatterns = [
url(r'^admin/', include(admin.site.urls)),
url(r'^xtrack/$', views.search_form),
url(r'^resultform/$', views.search),
url(r'^errorform/$', views.search)
]
and my templates are like: index.html
<html>
<h1> Welcome to xTrack </h1>
<head>
<title>Search</title>
</head>
<body>
<form action="/xtrack/" method="get">
<input type="text" name="xtrackid">
<input type="submit" value="Search">
</form>
</body>
</html>
resultform.html
Results
{% if contextresultinfo %}
<table border="1" style="width:100%">
<tr>
<td>xtide tracker ID<br> </td>
<td>Accession number<br></td>
</tr>
{% for key, values in contextresultinfo.items %}
<tr>
{% for items in values %}
<tr>
<td>{{key}}</td>
{% for data in items %}
<td>{{data}}</td>
{% endfor %}
</tr>
{% endfor %}
</tr>
{% endfor %}
</table>
{% else %}
<p>No xtrack matched your search criteria.</p>
{% endif %}
</body>
Can you please give some idea where do I need to change code in my project. Thanks
