javascript - Get data from mysql database using php and jquery ajax

I want to get data from mysql database using php and jquery ajax. 'process.php' is the php file which connects to database and get mysql data. It works when it is run separately, but when called using ajax it doesn't work. Can someone please help to correct error? Here is my html file:

<html>
<head>
<script type="text/javascript" src="js/jquery-1.11.0.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
    function showRoom(){
        $.ajax({
            type:"POST",
            url:"process.php",
            data:{action:showroom},
            success:function(data){
                $("#content").html(data);
            }
        });
    }
    showRoom();
});
</script>
</head>
<body>
<div id="content"></div>
</body>
</html>

Here is my process.php file

<?php
$link=mysqli_connect("localhost","root","raspberry","homebot");

if (mysqli_connect_errno())
    echo "Failed to connect to MySQL: " . mysqli_connect_error();

$action=$_POST["action"];
if($action=="showroom"){
    $query="SELECT * FROM user";
    $show=mysqli_query($link,$query) or die ("Error");
    while($row=mysqli_fetch_array($show)){
        echo "<li>$row['name']</li>";
    }
}
?>

<html>
<head>
<script type="text/javascript" src="js/jquery-1.11.0.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
    function showRoom(){
        $.ajax({
            type:"POST",
            url:"process.php",
            data:{action:showroom},
            success:function(data){
                $("#content").html(data);
            }
        });
    }
    showRoom();
});
</script>
</head>
<body>
<div id="content"></div>
</body>
</html>

Here is my process.php file

<?php
$link=mysqli_connect("localhost","root","raspberry","homebot");

if (mysqli_connect_errno())
    echo "Failed to connect to MySQL: " . mysqli_connect_error();

$action=$_POST["action"];
if($action=="showroom"){
    $query="SELECT * FROM user";
    $show=mysqli_query($link,$query) or die ("Error");
    while($row=mysqli_fetch_array($show)){
        echo "<li>$row['name']</li>";
    }
}
?>

Share Improve this question edited Mar 25, 2014 at 19:10 Jeff B 9,01220 gold badges67 silver badges144 bronze badges asked Mar 25, 2014 at 18:47 thomas 271 gold badge1 silver badge5 bronze badges

1 What doesn't work? Do you get an error? What is it? What debugging have you done? – John Conde Commented Mar 25, 2014 at 18:47
I dont get an error.But nothing is printed.Printing from mysql database is expected ($row[name]) – thomas Commented Mar 25, 2014 at 18:54
Add if(!mysqli_num_rows($show)) echo "no result" before while. Maybe the query doesn't return anything? Or try to put name in quotes. – Al.G. Commented Mar 25, 2014 at 19:01
added.but still same result – thomas Commented Mar 25, 2014 at 19:06
Don't use arrays in strings: replace echo "<li>$row['name']</li>"; with echo "<li>".$row['name']."</li>"; – Al.G. Commented Mar 25, 2014 at 19:13

Add a ment |

2 Answers 2

Sorted by: Reset to default 5

There are two syntax errors in your ajax call:

$(document).ready(function(){
    function showRoom(){
        $.ajax({
            type:"POST",
            url:"process.php",
            data:{action:"showroom"},
            success:function(data){
                $("#content").html(data);
            }
        });
    }
    showRoom();
});

Keep in mind that jQuery's ajax expects an object as parameter. Inside an object the syntax is

{ key : value }

You had type="POST" which is correct in declarative syntax, but incorrect when defining an object key.

Second, the data property of the aforementioned object should be an object too. So instead of action=showroom it should be

{action:"showroom"}

you did mistake in your code:

 echo "<li>$row['name']</li>";

This should be:

 echo "<li>".$row['name']."</li>";

try that...

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2 Answers 2

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