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Wordpress使用Wordpress将Ajax值传递到特定页面

ajax

我想将变量传递到特定页面。我找到了一个简单的示例,解释了如何在Wordpress中使用Ajax。

JavaScript:

jQuery(document).ready(function($) {

// We'll pass this variable to the PHP function example_ajax_request
var fruit = 'Banana';

// This does the ajax request
$.ajax({
    url: ajaxurl,
    data: {
        'action':'example_ajax_request',
        'fruit' : fruit
    },
    success:function(data) {
        // This outputs the result of the ajax request
        console.log(data);
    },
    error: function(errorThrown){
        console.log(errorThrown);
    }
});

});

插入的一段PHPfunctions.php

function example_ajax_request() {

// The $_REQUEST contains all the data sent via ajax
if ( isset($_REQUEST) ) {

    $fruit = $_REQUEST['fruit'];

    // Let's take the data that was sent and do something with it
    if ( $fruit == 'Banana' ) {
        $fruit = 'Apple';
    }

    // Now we'll return it to the javascript function
    // Anything outputted will be returned in the response
    echo $fruit;

    // If you're debugging, it might be useful to see what was sent in the $_REQUEST
    // print_r($_REQUEST);

}

// Always die in functions echoing ajax content
  die();
 }

add_action( 'wp_ajax_example_ajax_request', 'example_ajax_request' );


   wp_localize_script( 'ajax-script', 'ajax_object', array( 'ajax_url' =>   admin_url( 'admin-ajax.php' ) ) );

不幸的是我不能传递变量。我检查了代码,并得到以下错误:

Error: ajax_object is not defined

您是否知道另一种获得相同结果的方法?


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2020-07-26

共1个答案

小编典典

您距离很近,但缺少一些小东西……

我在评论中的意思是,您需要'ajax-script'在两种情况下都使用这种方式:

add_action('wp_enqueue_scripts', 'add_js_scripts'); 
add_js_scripts(){
    wp_enqueue_script( 'ajax-script', get_template_directory_uri().'/js/script.js', array('jquery'), '1.0', true );
    wp_localize_script( 'ajax-script', 'ajax_object', array( 'ajaxurl' =>   admin_url( 'admin-ajax.php' ) ) );
}

更改 $_REQUEST$_POST

function example_ajax_request() {

    // The $_REQUEST contains all the data sent via ajax
    if ( isset($_POST) ) {

        $fruit = $_POST['fruit'];

        // Let's take the data that was sent and do something with it
        if ( $fruit == 'Banana' ) {
            $fruit = 'Apple';
        }

        // Now we'll return it to the javascript function
        // Anything outputted will be returned in the response
        echo $fruit;

        // If you're debugging, it might be useful to see what was sent in the $_POST
        // print_r($_POST);

    }

    // Always die in functions echoing ajax content
      die();

 }

新增add_action( 'wp_ajax_nopriv_ … )

add_action( 'wp_ajax_nopriv_example_ajax_request', 'example_ajax_request' ); // <= this one
add_action( 'wp_ajax_example_ajax_request', 'example_ajax_request' );

对于您的jQuery脚本script.js文件,有2件重要的遗漏小事情:

jQuery(document).ready(function($) {

    /* We'll pass this variable to the PHP function example_ajax_request */
    var fruit = 'Banana';

    /* This does the ajax request */
    $.ajax({
        url: ajax_object.ajaxurl, /* <====== missing here */
        type : 'post', /*    <========== and missing here */
        data: {
            'action':'example_ajax_request',
            'fruit' : fruit
        },
        success:function(data) {
            /* This outputs the result of the ajax request */
            console.log(data);
        },
        error: function(errorThrown){
            console.log(errorThrown);
        }
    });

});

现在应该可以工作了……

参考文献:

2020-07-26