CODEWITHSUNDEEP
phpjquery
chao
chao

Reputation: 5

PHP get data from jquery

Im trying to write a simple prgram that the server can get data from client. I write a simple code in my script

var str = "testString"; 

$.post("http://anonymous.comze.com/test1.php", { string: str });

in the server,

$var = $_POST['string']; // this fetches your post action

$sql2 = "INSERT INTO afb_comments VALUES ('3',$var)";

$result2= mysql_query($sql2,$conn);

The question is var is always null. The sql2 can be executed if I change $var into "1111" for example, but if I put $var, it doesn't work. Can anyone give some advice?

Upvotes: 0

Views: 84

Answers (2)

Walter Jamison
Walter Jamison

Reputation: 36

Use this example and learn from this code how to get data

Or

use can also use this link:

http://api.jquery.com/jQuery.get/

$user and $pass should be set to your MySql User's username and password.

I'd use something like this:

JS

success: function(data){
             if(data.status === 1){
                 sr = data.rows;
             }else{
                 // db query failed, use data.message to get error message
             }
        }
PHP:

<?php

    $host = "localhost";
    $user = "username";
    $pass = "password";
    $databaseName = "movedb";
    $tableName = "part parameters";

    $con = mysql_pconnect($host, $user, $pass);
    $dbs = mysql_select_db($databaseName, $con);
    //get the parameter from URL
    $pid = $_GET["pid"];
    if(empty($pid)){
        echo json_encode(array('status' => 0, 'message' => 'PID invalid.'));
    } else{
        if (!$dbs){
            echo json_encode(array('status' => 0, 'message' => 'Couldn\'t connect to the db'));       
        }
        else{
            //connection successful
            $sql = "SELECT `Processing Rate (ppm)` FROM `part parameters` WHERE `Part Number` LIKE `" . mysqli_real_escape_string($pid) . "`"; //sql string command
            $result = mysql_query($sql) or die(mysql_error());//execute SQL string command
            if(mysql_num_rows($result) > 0){
                $rows = mysql_fetch_row($result);
                echo json_encode(array('status' => 1, 'rows' => $rows["Processing Rate (ppm)"]);
            }else{
                echo json_encode(array('status' => 0, 'message' => 'Couldn\'t find processing rate for the give PID.'));   
            }
        }

    }

?>
As another user said, you should try renaming your database fields without spaces so part parameters => part_parameters, Part Number => part_number.

Upvotes: 0

Pragnesh Chauhan
Pragnesh Chauhan

Reputation: 8476

your are passing string to the query so it should be

$var = $_POST['string']; // this fetches your post action

$sql2 = "INSERT INTO afb_comments VALUES ('3','".$var."')";

$result2= mysql_query($sql2,$conn);

please also check datatype of the that column.

Upvotes: 1

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