Question 37231 in Jquery Question posted in
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PHP doesn't display database data, but displays echo from the same function – Jquery

I’m using AJAX to transport id name of my div to use it as a string in a query. The problem is that it does not want to display any content related to the database (it can display a normal text echo from my function).

// head HTML (AJAX)
$(function () { 
        $('.product').on('click', function (e) {
            var product = $(this).attr('id');
            
            $.ajax({
                type: "post",
                url: 'php/recipe-container.php',
                data: JSON.stringify({
                    data: product,
                }),
                processData:false,
                processData:false,
                contentType: false,
                success: function(response) {
                    $(".display_recipe").html(response);
                }
            });
        });
});

//HTML
<div class="product" id="pizza">pizza</div>
<div class="product" id="lasagna">lasagna</div>
<div class="product" id="sushi">sushi</div>
<div class="display_recipe">
  <?php
    require "php/recipe-container.php";
  ?>
</div>

//PHP (recipe-container.php)
<?php

function display(){
    $con = mysqli_connect("localhost", "root", "", "cookbook");
    echo "test ";
    $product = substr(file_get_contents("php://input"), 9, -2);
    
    $sql = "SELECT * FROM product WHERE name = '$product'";
    $res = mysqli_query($con,$sql);

    while($row = mysqli_fetch_assoc($res)) {
        echo "test2 ";
        $name = $row['name'];
        $description = $row['description'];
        $date = $row['date'];

        echo $name;
        echo $description;
        echo $date;
    }

mysqli_close($con);
}

if ($_SERVER['REQUEST_METHOD'] === 'POST'){ 
    display(); 
}

?>

This is what I see after the button click – "test test2 pizza". "pizza" is the ID name, not the actual name from the database. I don’t know how to remove that. I know everything has something to do with the "success" function (response), but I have no idea how to fix that.

Tags:

2

Answers


  1. 0

    Your solution is:

    Change:

    var product = $(this).attr('id');

    To:

    var product = $(this).html();
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  2. 0

    you have to make an array in php before while loop as:

    $data = [];
while($row = mysqli_fetch_assoc($res)){

    $data['name'] = $row['name'];
    $data['description'] = $row['description'];
}
return $data;
// return json_encode($data);
exit();

    I hope this works fine for you. Now you just have to pass json format in your ajax request and decode and then it works fine for you

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