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Editorial Team
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Asked: 2026-05-17T19:45:21+00:00

I have a php file that receives data from mySQL table. The mySQL table

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I have a php file that receives data from mySQL table. The mySQL table ‘user_spec’ has only one field ‘options’ that it returns. Then I convert returned data into JSON, under is code doing that.

<?php 
 $username = "user"; 
 $password = "********"; 
 $hostname = "localhost"; 
 $dbh = mysql_connect($hostname, $username, $password) or die("Unable to connect  
         to MySQL"); //print "Connected to MySQL<br>"; 
 $selected = mysql_select_db("spec",$dbh) or die("Could not select first_test"); 
 $query = "SELECT * FROM user_spec"; 
 $result=mysql_query($query);  
 echo json_encode(mysql_fetch_assoc($result)); 
?>

then in an HTML file, I try to output data by this piece of code But it is not working. I will be very thankful for any help. 

<html>
    <head>
    <script type="text/javascript"   
    src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js"></script>
    <script type="text/javascript" language="javascript">
    function Preload() {
    $.getJSON("Dhttp://localhost/conn_mysql.php", function(json){
    alert("JSON Data: " + json.user_spec);
    });}

    </script></head>
    <body onLoad="Preload()">
    </body>
    </html>
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1 Answer

  1. Editorial Team

    I think instead of using the file adress: “D:xampp/htdocs/conn_mysql.php” you must use an url defined by xampp such as “http://localhost/mytest/conn_mysql.php”

    Another thing you need to look out is the method. $.getJSON (http://api.jquery.com/jQuery.getJSON/), as its name says, works with the GET method. Maybe you should try $.post or $.ajax (http://api.jquery.com/jQuery.post/).

    Oh! And to start your script, not all the browsers supports the <body onload=””>. Furthermore, it waits for the page to be loaded, not the DOM, what can give you problems with your scripts sometimes.
    You should use $(‘document’).ready(function(), that waits for the DOM to be loaded.
    This way:

    <script type="text/javascript">
$('document').ready(function()
{

    $.getJSON("D:xampp/htdocs/conn_mysql.php", function(json){
    alert("JSON Data: " + json.options);

});
</script>

    I hope it can be useful! ^^

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