Question

我创建了一个代码，允许您从数据库中搜索视频。但是每当我写下我要搜索的视频时，在控制台中出现error ......我真的不知道为什么它不打印结果......

你能帮帮我吗？

我需要快速帮助

AJAX

$(document).ready(function($) {
var Search = function(title){ return $.post( "./include/php/busqueda_videos.php", { "title" : title }); }

 $("#input_search").on('keyup',  function(event) {
      var output = "";
    $(".videos_container").html("");
    event.preventDefault();
    var input = $("#input_search").val();
    Search(input).done(function(response) {
        if(response.success) {
                $.each(response.videos, function(key, value) { var output = "<div class='video_main_container'><div class='video_container'><div class='video_thumb'><a data-id='"+value['id']+"'><img src='"+value['image']+"' alt='' /></a></div><div class='video_info'><p class='title'>"+value['title']+"</p><p class='usuario'>"+value['user']+"</p></div></div></div>"; });
             $(".videos_container").html(output);   
           } else {
            console.log("Error");
        }
    }).fail(function(jqXHR, textStatus, errorThrown) { console.log("Hubo un error"); });
 });

});

PHP

<?php
if(isset($_POST['title'])) {
get_infvideo($_POST['title']);
} else {
$message = sprintf("No valid");
header($_SERVER['SERVER_PROTOCOL'] . ' ' . $message, true, 403);
}


function get_infvideo($title) {
$dsn = "mysql:host=localhost;dbname=tapehd;charset=utf8";
$usuario = "root";
$contraseña = "";

$conexion = new PDO($dsn, $usuario, $contraseña);
$resultado = null;
$jsondata = array();
$videos = array();
$text ="";
if($title!="") {
    $title= explode(" ", $title);
    for($i=0; $i<count($title);$i++) {
        if($text!="") {
            $text .= " AND (title LIKE '%".$title[$i]."%' or user LIKE '%".$title[$i]."%')";
        } else {
            $text .= " (titulo LIKE '%".$titulo[$i]."%' or user LIKE '%".$title[$i]."%')";
        }
    }
}

if($conexion){
    $sql = "SELECT * FROM video WHERE ".$text;
    if($resultado = $conexion->query($sql)) {
        while($fila = $resultado->fetch()) {    
            $fila['id'] =  $fila['id'];
            $fila['user'] = $fila['user'];
            $fila['image'] =  $fila['image'];
            $jsondata['success'] = true;
            array_push($videos, $fila);
            $jsondata['videos'] = $videos;
        }
    } else {
        $jsondata['success'] = false;
        $jsondata['message'] = $sql;
    }
}

header('Content-type: application/json; charset=utf-8');
echo json_encode($jsondata, JSON_FORCE_OBJECT);
}

exit();
?>

我们的想法是打印您搜索过的视频（包括他的ID，图片，用户，标题）......

尝试搜索视频时遇到错误AJAX

0 个答案: