jQuery 调用 PHP 文件从 mysql 数据库中获取数据？

Question

Ok so I have found a number of tutorials online and have followed each of them step by step. My problem is that I know javascript/jQuery much better than I know PHP and I cannot figure out how to even debug what is going wrong in that section. Basically I have a bunch of buttons and a from and when a button is pressed it determines what the default values are in the form.

jQuery Side

$(document).ready(function(){
// CSPC and ADDUPDATE TOGGLE ARE DEFINED GLOBALLY
    $('ul#parts').on('click', 'button', function(){
        ADDUPDATETOGGLE = "ADD";
        CSPC = $(this).attr("data-cspc");
        var   form = $('div.sidebar form'),
              sr = 0;
        form.find("#cspc").val(CSPC);
        $.ajax({
            type: "GET",
            url: "getRate.php",
            data: "pid=A5843",
            dataType: "json",
            success: function(data){
                sr = data;
            }
        });
        form.find("#strokeRate").val(sr);
        showForm();
    });
});

PHP side

<?php

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

$con = mysql_connect($host, $user, $pass);
$dbs = mysql_select_db($databaseName, $con);
//get the parameter from URL
$pid=$_GET["pid"];
if (empty($pid)){
    echo "1"; //default rate
}
else{
    $db=mysql_pconnect("localhost");//connect to local database
    mysql_select_db("movedb", $db);//select the database you want to use
    if (!$db){
        echo ("error connecting to database");              
    }
    else{
        //connection successful
        $sql = "SELECT 'Processing Rate (ppm)' FROM 'part parameters' WHERE 'Part Number' LIKE '" . $pid . "'";//sql string command
          $result=mysql_query($sql);//execute SQL string command
          //result contains rows
          $rows = mysql_fetch_row($result)
          echo json_encode($rows["Processing Rate (ppm)"]); 
    }

}

?>

Any ideas why sr is not getting set?

Am I way off base?

I will also shamelessly note that I do not know what $user and $pass should be set to. I cannot find that explained anywhere

Thanks in advance

EDIT: I followed most of the directions below and now when I run

http://localhost/getRate.php?pid=A5843 

it says "No database selected." Also, I dont have access to our original MS Access file now (one of my team members has it) but once I get it I will make all the headers into one word headers. This is our first job with web programming/database management so we are constantly learning.

Answer 1

$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 .

If you're still having trouble then (as long as it's not a production server) put this at the top of your php file:

error_reporting(E_ALL);
ini_set('display_errors', '1');

This will output any errors and should help you work out what's going wrong.

Answer 2

Your DB query code is incorrect:

    $sql = "SELECT 'Processing Rate (ppm)' FROM 'part parameters' WHERE 'Part Number' LIKE '" . $pid . "'";//sql string command

using ' to quote things in the query turns them into STRINGS, not field/table names. So your query is syntactically and logically wrong. Your code is simply assuming success, and never catches the errors that mysql will be spitting out.

The query should be:

SELECT `Processing Rate (ppm)`
FROM `part parameters`
WHERE `Part Number` = '$pid'

Note the use of backticks (`) on the field/table names, and the use of single quotes ( ' ) on the $pid value.

Then you execute the query with:

$result = mysql_query($sql) or die(mysql_error());

If this fails, you will get the error message that mysql returns.

And in the grand scheme of things, your code is vulnerable to SQL injection attacks . Better read up and learn how to prevent that before you go any farther with this code.

Answer 3

sr it's outside the success callback function. Start putting it into the success function and see what happens

$.ajax({
        type: "GET",
        url: "getRate.php",
        data: "pid=A5843",
        dataType: "json",
        success: function(data){
            sr = data;
            form.find("#strokeRate").val(sr);
        }
    });

remember that, if data is expected to be a json, it will become a js object, so you will not be able to use data directly

Answer 4

Here is a compilation of the above with up-to-date code.

jQuery

$(document).ready(function(){
...
    $.ajax({
            type: "GET",
            url: "getRate.php",  //see note "url"
            data: {
                    pid : "A5843" 
                 //, ... : "..." 
                  }, 
            dataType: "json",
            success: function(data){
                sr = data;
            }
        });
...
    });

• url > test your request-URL including any parameters (eg. http://localhost/XYZ/getRate.php?pid=251) and then enter it here, after removing anything after the '?' symbol (including '?')
• data > https://api.jquery.com/Jquery.ajax/

PHP

<?php

    $host = "localhost";
    $user = "username";
    $pass = "password";
    $databaseName = "movedb";
    
    $pid=$_GET['pid']; //get the parameter from URL

    $con = mysqli_connect($host, $user, $pass, $databaseName);
    //$dbs = mysqli_select_db($con, $databaseName);


    if (empty($pid)){
        echo json_encode(array('status' => 0, 'message' => 'pid invalid.'));
    }
    else{
        if (mysqli_connect_errno()) {
          echo "Failed to connect to MySQL: " . mysqli_connect_error();
          exit;
        }
        //if (!$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) . "`"; //https://www.geeksforgeeks.org/how-to-prevent-sql-injection-in-php/      
            $result = mysqli_query($con, $sql) or die(mysql_error());//execute query
            if($result){
                $rows = mysqli_fetch_row($result);
                echo json_encode(array('status' => 1, 'rows' => $rows["Processing Rate (ppm)"]);
            }
            else
                echo json_encode(array('status' => 0, 'message' => 'Couldn\'t find results'));   
            }
            mysqli_free_result($result);
            mysqli_close($con);
        }


?>

Please note that I am no PHP-expert. Any comments welcome.