AJAX + PHP Data Search

Question

I am using the code below to to pull information from my database, in the option menu, the data shows but for some reason the value is not being passed on to pull information using ajax. Please advise.

PHP:

 <html>
 <head>
 <script type="text/javascript">
 function showUser(str)
 {
 if (str=="")
 {
 document.getElementById("txtHint").innerHTML="";
  return;
  }
 if (window.XMLHttpRequest)
 {// code for IE7+, Firefox, Chrome, Opera, Safari
 xmlhttp=new XMLHttpRequest();
 }
 else
 {// code for IE6, IE5
 xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
 }
 xmlhttp.onreadystatechange=function()
 {
 if (xmlhttp.readyState==4 && xmlhttp.status==200)
 {
 document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
  }
  }
 xmlhttp.open("GET","ajax.php?q="+str,true);
  xmlhttp.send();
  }
  </script>
 </head>
 <body>
<select onchange="display_data(this.value);">
<option>Select user</option>
<?php
include("config.php");

$query="SELECT p_name, full_name FROM users order by p_name asc";
$result=mysql_query($query);
while(list($p_name, $full_name)=mysql_fetch_row($result)) {
    echo "<option value=\"".$p_name."\">".$full_name."</option>";
}

?>
</select>
<div id="txtHint"><div>
</body>
</html>

Ajax.php

 <?php
 $q=$_GET["q"];

 include("config.php");


 //Query
 $sql = "select `full_name` from `users` where `p_name` = '$q'";
 $query = mysql_query($sql) or die ("Error: ".mysql_error());

    while ($row = mysql_fetch_array($query)){

 $full_name = $row['Full_name'];


        print $full_name

        ?>

Hemant Jain · Accepted Answer · 2011-10-20 10:35:30Z

2

Jquery Ajax call through data serach from php file  

$('#btnSubmit').click(function() {  


   var name = $('#Name').val();

       $.ajax({  
           type: "POST",  
           url: "search.php",
           data: "name="+ name,  
           success: function(response){ 
            $('#search').html(response);
           }  
       });
  });

answered Oct 20, 2011 at 10:35

Hemant Jain

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Comments

CronosS · Accepted Answer · 2011-01-23 14:03:19Z

1

It seems that you don't close your while loop in your "ajax.php". The rest of the code seems to be ok.

answered Jan 23, 2011 at 14:03

CronosS

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4 Comments

AAA Over a year ago

you mean } after the $full_name?

CronosS Over a year ago

Yes. Php must rise a syntax error for now and not execute your code.

AAA Over a year ago

Did it but its still not working. Guess i'll never figure this ajax code out!

CronosS Over a year ago

There's also a semicolon missing after your print. Ma fault, I didn't notice before. Anyway, your ajax code works. The prob clearly comes from ajax.php.

user514310 · Accepted Answer · 2011-01-23 15:46:39Z

Now its working. There was two mistakes in you code. First one I correct onchange="showUser(this.value); and second one echo $row['full_name']; } Ajax.php

 <?php
$q=$_GET["q"];
include("config.php");
//Query
$sql = "select `full_name` from `users` where `p_name` = '$q'";
$query = mysql_query($sql) or die ("Error: ".mysql_error());
while ($row = mysql_fetch_array($query)){
    echo $row['full_name'];
}
?>

Index.php

 <html>
 <head>
 <script type="text/javascript">
 function showUser(str)
 {
if (str=="")
{
 document.getElementById("txtHint").innerHTML="";
return;
}
  if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
   {// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
 if (xmlhttp.readyState==4 && xmlhttp.status==200)
 {
 document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET","ajax.php?q="+str,true);
xmlhttp.send();
}
</script>
</head>
<body>
 <select onchange="showUser(this.value);">
 <option>Select user</option>
 <?php
 include("config.php");
$query="SELECT p_name, full_name FROM users order by p_name asc";
$result=mysql_query($query);
while(list($p_name, $full_name)=mysql_fetch_row($result)) {
echo "<option value=\"".$p_name."\">".$full_name."</option>";
}

?>
</select>
<div id="txtHint"><div>
</body>
</html>
enter code here

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