我如何将 jquery 值传递给 php 函数

Question

Here i want to call the php function with some parameter that will be held in some jquery function.

I have php function that takes one parameter and executes the sql query. and also i have a jquery function where i am getting different values when select item from dropdown box.

now i want to pass the value of dropdown box to php function, how can i do this? here is what i have done

$("#product_category").change(function(){
    var category = $(this).val(); //getting dropdown value
    // here i want to pass this variable value to php function
});

php

    function productManagement($procat){
        // Create connection
        $conn = new mysqli($servername, $username, $password, $dbname);
        // Check connection
        if ($conn->connect_error) {
            die("Connection failed: " . $conn->connect_error);
        }
    $sql = "SELECT product_name,product_category,product_image_url FROM product_list where product_category='$procat'";
    $result = $conn->query($sql);

    if ($result->num_rows > 0) {
        // output data of each row
        while($row = $result->fetch_assoc()) {
        echo "<tr><td>". $row["product_name"]."</td><td><a href=''><span class='glyphicon glyphicon-remove'></span>Remove</a></td><td><a href='edit-product.php'><span class='glyphicon glyphicon-edit'></span>Edit</a></td></tr>";
        }
    } else {
        echo "No results found";
    }
    $conn->close();
}

How can i do this?

Answer 1

Another way is to use Ajax like :

Jquery code:

$("#product_category").change(function(){
    var category = $(this).val(); //getting dropdown value
    // here i want to pass this variable value to php function

-----------------------------------------------------
$.ajax({
  url: 'newFunctionFile.php',
  type: 'POST',
  data: 'category='+category,
  success: function(data) {
    //you can add the code to show success message

  },
  error: function(e) {
    //called when there is an error
    //console.log(e.message);
  }
});

});

and you have to create a file for php function let say file name is newFunctionFile.php as i have mention in ajax call:

if(isset($_POST['category'])) {                
    productManagement($_POST['category']);    
}

 function productManagement($procat){
        // Create connection
        $conn = new mysqli($servername, $username, $password, $dbname);
        // Check connection
        if ($conn->connect_error) {
            die("Connection failed: " . $conn->connect_error);
        }
    $sql = "SELECT product_name,product_category,product_image_url FROM product_list where product_category='$procat'";
    $result = $conn->query($sql);

    if ($result->num_rows > 0) {
        // output data of each row
        while($row = $result->fetch_assoc()) {
        echo "<tr><td>". $row["product_name"]."</td><td><a href=''><span class='glyphicon glyphicon-remove'></span>Remove</a></td><td><a href='edit-product.php'><span class='glyphicon glyphicon-edit'></span>Edit</a></td></tr>";
        }
    } else {
        echo "No results found";
    }
    $conn->close();
}

Answer 2

How about something along these lines.

In your javascript:

$("#product_category").change(function(){
var category = $(this).val();                        //getting dropdown value
location.href = location.href + "?val="+category;    //reload the page with a parameter
});

And then in your PHP, add this code as your function call

if(isset($_GET['val'])) {                //if 'val' parameter is set (i.e. is in the url)
    productManagement($_GET['val']);     //call function, with that parameter
}