我正在将变量接收到JavaScript中，但没有在PHP中获取它，我认为它在JavaScript中的编码不正确：
function glitch_player_display(mix_name) {
alert(mix_name);
throb.start();
jQuery.ajax({
type: \'POST\',
url: ajaxglitch_playerajax.ajaxurl,
data: {
action: \'ajaxglitch_player_ajaxhandler\',
mix_name: mix_name
},
success: function(data, textStatus, XMLHttpRequest) {
throb.stop();
var showglitchplayer = \'#showglitchplayer\';
jQuery(showglitchplayer).html(\'\');
jQuery(showglitchplayer).append(data);
},
error: function(MLHttpRequest, textStatus, errorThrown) {
alert(errorThrown);
}
});
}
这是PHP：
add_action( \'wp_ajax_nopriv_ajaxglitch_player_ajaxhandler\', \'ajaxglitch_player_ajaxhandler\' );
add_action( \'wp_ajax_ajaxglitch_player_ajaxhandler\', \'ajaxglitch_player_ajaxhandler\' );
function ajaxglitch_player_ajaxhandler($arguments){
var_dump($arguments);
}
返回空。
谢谢，祝你过得愉快。
最合适的回答,由SO网友:doublesharp 整理而成
AJAX回调不接受任何参数，您必须使用$_GET/$_POST/$_REQUEST （取决于javascript的设置方式）。这意味着mix_name 从您的POST 您将使用$_POST[\'mix_name\']. 包括add_action( \'wp_ajax_... 如果您将来还有问题的话，从代码中调用会很好，我不得不去GitHub页面查看它是如何设置的-您还缺少单引号add_action( \'wp_enqueue_scripts\', \'ajaxglitch_player_enqueuescripts\' ); 哪一个PHP可以弄清楚，但它会先将其作为常量进行尝试。配置应该是这样的：
function ajaxglitch_player_enqueuescripts() {
wp_enqueue_script(\'ajaxglitch_player\', glitch_player_URL.\'/js/glitch_player.js\', array(\'jquery\'));
// access ajax url on the front end as ajaxglitch_playerajax.ajaxurl
wp_localize_script( \'ajaxglitch_player\', \'ajaxglitch_playerajax\', array( \'ajaxurl\' => admin_url( \'admin-ajax.php\' ) ) );
}
add_action( \'wp_enqueue_scripts\', \'ajaxglitch_player_enqueuescripts\' );
// front end ajax handler
add_action( \'wp_ajax_nopriv_ajaxglitch_player_ajaxhandler\', \'ajaxglitch_player_ajaxhandler\' );
// admin ajax handler
add_action( \'wp_ajax_ajaxglitch_player_ajaxhandler\', \'ajaxglitch_player_ajaxhandler\' );
function ajaxglitch_player_ajaxhandler(){
$mix_name = isset( $_POST[\'mix_name\'] )? $_POST[\'mix_name\'] : false;
var_dump( "The mix_name is $mix_name" ); // this will probably break the AJAX response
error_log( "The mix_name is $mix_name" ); // write it to the error_log too.
}
