Adarsh Sharma am Donnerstag, 30. Januar 2014 - 12:25: > 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 > > > I'm sorry, I cannot test your site: DNS failure! host reports: "Host db11.app.com not found: 3(NXDOMAIN)" If ist works for You, the error must be on the ajax side. Personaly I was never using ajax. So: refer to the ajax Manual :-) Thanks, Niklaus -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
