The one thing i doubts whether program control reaches to ajaxex.php.Access_log says it is executing ajaxex.php. I tried to check whether my program flow reaches to ajaxex.php by adding below lines in ajaxex.php : //echo "Mysql Connected"; //echo "Value of users : $q"; before the select command but after clicking on Weekly Report nothing happens :( Seems debugging is very difficult than to fix. There is no error in error_log as well. Thanks On Thu, Jan 30, 2014 at 4:55 PM, Adarsh Sharma <eddy.adarsh@xxxxxxxxx>wrote: > There is no error in mysql queries i.e i told earlier if u opens the > below link directly : > > *http://db11.app.com/project/ajaxex.php?q=133 > <http://db11.app.com/project/ajaxex.php?q=133>* > > you will get the complete report so no issues with ajaxex.php. > > but i need the same after clicking on Weekly Report Command button when i > opened > > > > *http://db11.app.com/project > <http://db11.app.com/project/ajaxex.php?q=133> and chooses one user. * > > *Thanks* > > > On Thu, Jan 30, 2014 at 4:45 PM, Domain nikha.org <mail@xxxxxxxxx> wrote: > >> Adarsh Sharma am Donnerstag, 30. Januar 2014 - 11:36: >> > PFA >> > >> > Thanks >> > >> > >> > On Thu, Jan 30, 2014 at 3:57 PM, Adarsh Sharma <eddy.adarsh@xxxxxxxxx >> >wrote: >> > >> > > Agreed Nikha. Valid point. See i tested by added one line in my code : >> > > >> > > >> > > var users = document.getElementById('users').value; >> > > var queryString = "?users=" + users ; >> > > * document.write(users); * >> > > ajaxRequest.open("GET","ajaxex.php?q="+users,true); >> > > ajaxRequest.send(); >> > > >> > > When i run it , it gives me exact value of the user which i chooses. >> If my >> > > user variable is empty then my output should be blank. Also, >> > > >> > > http:///db11.app.com/project/ajaxex.php?q=133 >> > > >> > > It the end of the above link , you can see the exact value of user. >> > > >> > > Thanks >> > > >> > > >> > > On Thu, Jan 30, 2014 at 3:50 PM, Domain nikha.org <mail@xxxxxxxxx> >> wrote: >> > > >> > >> Adarsh Sharma am Donnerstag, 30. Januar 2014 - 10:39: >> > >> > Thanks Nikha but its already dere : >> > >> > >> > >> > <html> >> > >> > <head> >> > >> > <script> >> > >> > function week_report_fun(){ >> > >> > var ajaxRequest; // The variable that makes Ajax possible! >> > >> > >> > >> > try{ >> > >> > // Opera 8.0+, Firefox, Safari >> > >> > ajaxRequest = new XMLHttpRequest(); >> > >> > }catch (e){ >> > >> > // Internet Explorer Browsers >> > >> > try{ >> > >> > ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP"); >> > >> > }catch (e) { >> > >> > try{ >> > >> > ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP"); >> > >> > }catch (e){ >> > >> > // Something went wrong >> > >> > alert("Your browser broke!"); >> > >> > return false; >> > >> > } >> > >> > } >> > >> > } >> > >> > ajaxRequest.onreadystatechange = function(){ >> > >> > if(ajaxRequest.readyState == 4){ >> > >> > var ajaxDisplay = document.getElementById('ajaxDiv'); >> > >> > ajaxDisplay.innerHTML = ajaxRequest.responseText; >> > >> > } >> > >> > } >> > >> > var users = document.getElementById('users').value; >> > >> > var queryString = "?users=" + users ; >> > >> > //document.write(queryString); >> > >> > ajaxRequest.open("GET","ajaxex.php?q="+users,true); >> > >> > ajaxRequest.send(); >> > >> > } >> > >> > >> > >> > </script> >> > >> > </head> >> > >> > <body> >> > >> > >> > >> > <form> >> > >> > User : <select id="users"> >> > >> > <option value="">Select a user</option> >> > >> > <option value="133">a.v</option> >> > >> > <option value="168">ad</option> >> > >> > </select> >> > >> > <input type='button' onclick="week_report_fun()" >> > >> > value='Weekly Report'/> >> > >> > </form> >> > >> > <br> >> > >> > >> > >> > </body> >> > >> > </html> >> > >> > >> > >> > >> > >> > Sorry My bad , i didn't mention the complete program in my first >> mail >> > >> > >> > >> > >> > >> > Thanks >> > >> > >> > >> > On Thu, Jan 30, 2014 at 2:36 PM, Domain nikha.org <mail@xxxxxxxxx> >> > >> wrote: >> > >> > >> > >> > > Adarsh Sharma am Donnerstag, 30. Januar 2014 - 07:27: >> > >> > > >> > >> > > > Hi, >> > >> > > > >> > >> > > > I am using AJAX and php for creating a small UI that retrieves >> data >> > >> from >> > >> > > > mysql upon clicking on one button. But i clicked on that , >> nothing >> > >> is >> > >> > > > happening. Below is my index.html : >> > >> > > > >> > >> > > > *-- index.html * >> > >> > > > >> > >> > > > <html> >> > >> > > > <head> >> > >> > > > function week_report_fun(){ >> > >> > > > var ajaxRequest; >> > >> > > > --- >> > >> > > > --- >> > >> > > > -- >> > >> > > > ajaxRequest.open("GET","ajaxex.php?q="+users,true); >> > >> > > > ajaxRequest.send(); >> > >> > > > } >> > >> > > > >> > >> > > > </script></head><body> >> > >> > > > <form> >> > >> > > > User : <select id="users" onchange="week_report_fun()"> >> > >> > > > <option value="">Select a user</option> >> > >> > > > <option value="133">av</option> >> > >> > > > <option value="168">ad</option> >> > >> > > > <option value="78">so</option> >> > >> > > > <option value="239">ak</option> >> > >> > > > </select></form><br> >> > >> > > > <div id="txtHint"><b>Person info will be listed here.</b></div> >> > >> > > > </body></html> >> > >> > > > >> > >> > > > *Access Log Output :-* >> > >> > > > >> > >> > > > 1.4.15.25 - - [30/Jan/2014:06:05:52 +0000] "GET >> > >> /project/ajaxex.php?q=133 >> > >> > > > HTTP/1.1" 200 5323 "http://db11.app.com/project/"; "Mozilla/5.0 >> > >> > > (Macintosh; >> > >> > > > Intel Mac OS X 10_9_1) AppleWebKit/537.36 (KHTML, like Gecko) >> > >> > > > Chrome/32.0.1700.102 Safari/537.36" >> > >> > > > 1.4.15.25 - - [30/Jan/2014:06:06:13 +0000] "GET >> > >> > > /project/ajaxex.php?q=133 >> > >> > > > HTTP/1.1" 200 5323 "-" "Mozilla/5.0 (Macintosh; Intel Mac OS X >> > >> 10_9_1) >> > >> > > > AppleWebKit/537.36 (KHTML, like Gecko) Chrome/32.0.1700.102 >> > >> > > Safari/537.36" >> > >> > > > >> > >> > > > No errors in error.log. >> > >> > > > >> > >> > > > Strange thing if i directly open below link in the browser, >> report >> > >> is >> > >> > > > coming and i am able to see in the browser : >> > >> > > > >> > >> > > > *http:///db11.app.com/project/ajaxex.php?q=133 >> > >> > > > <http://db11.app.com/project/ajaxex.php?q=133>* >> > >> > > > >> > >> > > > It means my ajaxex.php is fine and no issues but dont >> understand >> > >> why it >> > >> > > is >> > >> > > > not working as expected after clicking on button.Please let me >> know >> > >> if i >> > >> > > am >> > >> > > > missing something. >> > >> > > > >> > >> > > > >> > >> > > > Thanks >> > >> > > > >> > >> > > Hi Adarsh! >> > >> > > Yes, you are missing something: The opening <script> tag in your >> html >> > >> head! >> > >> > > without that, your ajax code is simply ignored. >> > >> > > >> > >> > > Hope it helps >> > >> > > Niklaus >> > >> > > >> > >> > >> > >> OK! >> > >> Consider to give names to the <form> and the <select> tag. >> > >> FI: >> > >> <form name="selector"> >> > >> <select name="user"> >> > >> >> > >> Then you can point to them in your javascript like this: >> > >> var users = document.selector.user.value >> > >> >> > >> Actualy, your users variable is just empty, what is not an error. So >> you >> > >> have no error reported. >> > >> >> > >> Niklaus >> > >> >> >> Hard case, I think! >> Thanks for your PHP script. What happens, if you run it without the >> preliminary javascript UI? >> >> The database query looks very complicated. I'm not realy an expert in >> mysqli. But as far as I experienced that, "SELECT" expects a _comma >> separated_ list of valid expressions before the "FROM" keyword. You have >> blanks there, like "us.login user" and others. Are you shure about this? >> >> mysqli_query() does not return error messages, so you have blanc pages if >> it failes, >> >> I think, you should have a look to the Mysql manual... >> Niklaus >> > >
