使用Ajax在表中显示PHP脚本中的数据

Question

I am looking to display data by running some query via php script and then using ajax to show it on an html page.

I have a php script that echos the data from a sql query in json format. The output looks like this:

{"Username":"Server","RICS":12739,"Exclusive_RICS":0}{"Username":"eikon1","RICS":4,"Exclusive_RICS":0}{"Username":"balbla","RICS":552,"Exclusive_RICS":0}{"Username":"somename","RICS":221,"Exclusive_RICS":201}

I would like to display this data using an $.ajax call.

I did some research and came up with this:

$(document).ready(function(){

$.ajax({

    url : 'query.php',
    type : 'POST',
    data : {
        //'numberOfWords' : 10
    },
    dataType:'json',
    success : function(data) {              
        window.alert(data.Username)
    },
    error : function(request,error)
    {
        alert("Request: "+JSON.stringify(request));
    }
});

});

However, it's not working properly. I just get this alert:

I am new to js/ajax/php so excuse me if I missed something simple.

Any help is appreciated. Thanks!

EDIT:

php code:

    $sql = 'select * from table';

$retval = sqlsrv_query($conn,$sql);
if(! $retval )
{
  die('Could not get data: ' . mysql_error());
}


while( $row = sqlsrv_fetch_array( $retval, SQLSRV_FETCH_ASSOC) ) {
    echo json_encode($row);
}

sqlsrv_free_stmt($retval);
//echo "Fetched data successfully\n";
sqlsrv_close($conn);

EDIT 2: Managed to get the output of php in correct JSON format through this. Now just need to display this using ajax.

while( $row = sqlsrv_fetch_array( $retval, SQLSRV_FETCH_ASSOC) ) {
    $rows[] = $row;
}
echo json_encode($rows);

Answer 1

Looping through your SQL result set and echoing each row separately produces an invalid JSON format. Instead you should json_decode the entire SQL result set.

Here's how you can update your PHP code so that it outputs the correct format:

php code:

$sql = 'select * from table';

$retval = sqlsrv_query($conn,$sql);
if(! $retval ) {
  die('Could not get data: ' . mysql_error());
}

echo json_encode( sqlsrv_fetch_array( $retval, SQLSRV_FETCH_ASSOC) );

sqlsrv_free_stmt($retval);
//echo "Fetched data successfully\n";
sqlsrv_close($conn);

There may be one more step necessary in your AJAX success callback. You'll have to JSON.stringify the result because PHP will send it back as plain text:

success : function(data) {              
    var result = JSON.stringify( data );
    window.alert(result.Username)
}, 

Answer 2

Thanks for the help everyone. I was eventually able to figure out the issue.

• First of all, as pointed by users, my json format was not right. I fixed that with the solution in my edit.
• Secondly, I had to reference my php directly with the exact address. I think it has to do with running queries from same server. Not sure perfectly.
• Thirdly, i tried a plain ajax call and even that was failing. It turns out my browser (chrome) needed a clean up. I cleared my cookies and it started to work fine. Why it was acting weird? I have no idea!

• Finally, now I needed to display the data in a table format and update the table frequently. I am still working with that but this is what I got going for me right now:

   $(document).ready(function() { (function poll() { $.ajax({ url : 'http://localhost/scripts/query.php', type : 'POST', data : {}, dataType:'json', success : function(response) { var json_obj = $.parseJSON(JSON.stringify(response)); var output=""; for (var i in json_obj) { output+="<tr>"; output+="<td>" + json_obj[i].time.date + "</td>" + "<td>" + json_obj[i].username + "</td>" + "<td>" + json_obj[i].rics + "</td>" + "<td>" + json_obj[i].exclusive_rics +"</td>"; output+="</tr>"; } $('#table_content').html(output); }, error : function(request,error) { alert("Request: "+JSON.stringify(request)); } , dataType: "json", complete: setTimeout(function() {poll()}, 5000), timeout: 2000 }) })(); 

  });

I really don't understand why this was so hard to do. I am sure this is a common scenario and I really wish there was a more straightforward way of doing this. Hopefully, my final code will avoid others from wasting so much of their time. Good luck!