使用AJAX将SQL WHERE查询获取到javascript中

Question

I am trying to pass a javascript variable into an SQL WHERE query and I keep getting null in return.

On-click of a button, the buttonClick function is ran:

<script>
    var var1;
    function buttonClick(elem){
        var1 = elem.src              //this gets the url from the element
        var path = var1.slice(48);   //this cuts the url to img/art/9/1.jpg
        ajax = theAjax(path);
        ajax.done(processData);
        ajax.fail(function(){alert("Failure");});
    }

    function theAjax(path){
        return $.ajax({
            url: 'info.php',
            type: 'POST',
            data: {path: path},
        });
    }

    function processData(response_in){
        var response = JSON.parse(response_in);
        console.log(response);
    }
</script>

Here is the code stored in the info.php file:

<?php
    $path = $_POST['path'];

    $result3 = mysqli_query("SELECT itemName from images WHERE imgPath='$path'");
    $json = json_encode($result3);

    echo $json
?>

As you can see, once I click the button, the buttonClick() function is ran and a variable stores the image path or src. That path variable is send to theAjax() function where it is passed to the info.php page. In the info.php page, the SQL WHERE query is ran and returned to the processData() function to be parsed and printed in the developer console. The value printed shows null .

Below is a picture of what I am trying to get from the database:

Answer 1

1.Check that path is correct or not? you can check inside jquery using console.log(path); or at PHP end by using print_r($_POST['path']);

2.Your Php code missed connection object as well as record fetching code.

<?php

    if(isset($_POST['path'])){
        $path = $_POST['path'];

        $conn = mysqli_connect ('provide hostname here','provide username here','provide password here','provide dbname here') or die(mysqli_connect_error());

        $result3 = mysqli_query($conn,"SELECT itemName from images WHERE imgPath='$path'");

        $result = []; //create an array

        while($row = mysqli_fetch_assoc($result3)) {
             $result[] = $row; //assign records to array
        }
        $json = json_encode($result); //encode response
        echo $json; //send response to ajax
    }

?>

Note:- this PHP query code is wide-open for SQL INJECTION . So try to use prepared statements of mysqli_* Or PDO .

Answer 2

mysqli_query()需要第一个参数作为连接对象。

$result3 = mysqli_query($conn,"SELECT itemName from images WHERE imgPath='$path'"); // pass your connection object here

Answer 3

I think your issue is that you're trying to encode a database resource.

Try adjusting your PHP to look like the following:

<?php
    $path = $_POST['path'];

    $result3 = mysqli_query("SELECT itemName from images WHERE imgPath='$path'");
    $return_data = [];
    while($row = mysqli_fetch_assoc($result3)) {
        $return_data[] = $row;
    }
    $json = json_encode($return_data);

    echo $json
?>