INSERT数据来自追加脚本(2),输入类型文件

时间:2016-06-26 14:11:38

标签: javascript php

我有问题。这是我的php& HTML:

<?php
      if ($_POST['btn_tambah'] == 'tambah') {

      $pathGambar = "gallery/" . $_FILES['gambar']['name'];         
      $SQL = "SELECT AUTO_INCREMENT as IDLapangan FROM information_schema.tables WHERE TABLE_SCHEMA = 'ta' AND TABLE_NAME = 'lapangan';";
      $res = mysql_query($SQL, $link);
      $row = mysql_fetch_object($res);
      foreach ($pathGambar as $path => $value) {
          $tambahGambar = mysql_query("INSERT INTO foto(path,id_portfolio) VALUES('" . $value . "',$row->IDPortfolio);");
      }
?>

<HTML><BODY>
  <div class="row">
   <div class="form-group">
      <div class="col-lg-3"><label>Gambar :</label></div>
      <div class="col-lg-2">
         <input type="file" name="gambar[]" class="form-control"  required>
      </div>
      <div class="col-lg-1">
       <a onclick="tambahGambarBaru()" class ="btn btn-info"> <i class="fa fa-plus"></i></a>
      </div>
    </div>
   </div>
  <div id="sembunyiGambar">
  </div>
</BODY></HTML>

这是我的剧本:

 function tambahGambarBaru() {
            $('#sembunyiGambar').append(                
                '<div class="row" id="imgBaru">'
                + '<div class="form-group">'
                + '<div class="col-lg-3">'
                + '</div>'
                + '<div class="col-lg-4">'
                + ' <input id="filePic" accept="image/*" type="file" name="gambar[]" class="form-control"  required>'
                + '</div>'
                + '<div class="col-lg-1">'
                + '<a class ="btn btn-warning" onclick="hapus()"> <i class="fa fa-trash"></i></a>'
                + '</div>'
                + '</div>'
                + '</div>'
        )
        }
 function hapus() {
            $('#imgBaru').remove();
        }

这是我的照片:

enter image description here

所以场景是,当我点击&#34;加&#34;按钮,它将显示第二个文本文件输入。我想将它们插入数据库。但是当我尝试插入时,它显示错误

  

为foreach()提供的无效参数

要显示第二个文本文件输入,我在javascript中使用.append

1 个答案:

答案 0 :(得分:0)

对不起这是我的错。 所以,它应该是这样的:

$pathGambar = $_FILES['gambar']['name'];  
 foreach ($pathGambar as $path => $value) {
      $tambahGambar = mysql_query("INSERT INTO foto(path,id_portfolio) VALUES('" . $value . "',$row->IDPortfolio);");
  }
非常感谢:)