Re: PHP output fails from AJAX call
| From: | Adarsh Sharma | Date: | Thu, 30 Jan 2014 10:36:00 +0000 |
| Subject: | Re: PHP output fails from AJAX call | ||
| References: | 1 2 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-322773@lists.php.net to get a copy of this message | ||
PFA
Thanks
On Thu, Jan 30, 2014 at 3:57 PM, Adarsh Sharma <eddy.adarsh@gmail.com>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@nikha.org> 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@nikha.org>
>> 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
>>
>>
>>
>
Attachment: [application/x-httpd-php] ajaxex.php
Attachment: [application/x-httpd-php] ajaxex.php