表单操作不会重定向到一个php文件

Question

我通过在我的工作中克隆另一个网站来创建一个新网站，并且我没有创建或编码这个。

所以，我的问题是：该网站正在工作，但图像上传不起作用，我不知道发生了什么事情。我尝试了很多东西来解决这个问题，比如var_dump在变量中尝试调试。下面是代码：

<? require_once("header.php"); ?> 
<? require_once("menu.php"); ?> 
<? require_once("ConnFile.php"); ?> 

<? 

if ($_GET["acao"] == "excluir") 
{ 
    $Tabela = "BannerHome"; 
    $TabelaChave = "idBanner"; 

if(is_numeric($_GET["k"])) 
{ 
    $txtIdProduto = $_GET["k"]; 
    $acao = ("UPDATE $Tabela SET Imagem". $_GET["I"] . " = '' WHERE $TabelaChave = $txtIdProduto "); 

    if(mysqli_query($link, $acao)) { 
     echo '<div class="alert alert-success alert-dismissible"" role="alert"> <button type="button" class="close" data-dismiss="alert"><span aria-hidden="true">&times;</span><span class="sr-only">Close</span></button> Registro removido com sucesso</div>'; 
    } else { 
     echo '<div class="alert alert-danger alert-dismissible"" role="alert"> <button type="button" class="close" data-dismiss="alert"><span aria-hidden="true">&times;</span><span class="sr-only">Close</span></button> Não foi possível remover o registro. </div>'; 
    } 
} 
} 

?> 
<? 
$id = 0; 
if(is_numeric($_GET["k"])) { 
    $id = $_GET["k"]; 
} 

$DadosProduto = mysqli_fetch_array(mysqli_query($link, "SELECT * FROM BannerHome WHERE idBanner = " . $id)); 

?> 

<? if($DadosProduto["Imagem" . $_GET["I"]] == "") { print '<div class="alert alert-warning alert-dismissible"" role="alert"> <button type="button" class="close" data-dismiss="alert"><span aria-hidden="true">&times;</span><span class="sr-only">Close</span></button> Este banner ainda não possui imagem cadastrada. </div>'; } ?> 

<? 


    ?> 

<link href="upload/assets/css/style.css" rel="stylesheet" /> 
<h4>Banner home<small>/Imagems/<?=$DadosProduto["Titulo"]; ?> </small></h4> 


<form class="form-horizontal" role="form" id="upload" action="upload/uploadHome.php?k=<?=$id; ?>&I=<?=$_GET['I']; ?>" method="post" enctype="multipart/form-data"> 
    <div id="drop"> 
    <pre> 
    <center> 
      Arreste as imagens (Fundo 1920x400, Superior 420x40) 
      <br> 
      ou 
      <br> 
      <a class="btn btn-primary btn-lg" >Selecione as imagens</a> 
       <input type="file" name="upl" multiple /> 
      </div> 

    </center> 
    </pre> 
      <ul> 
       <!-- The file uploads will be shown here --> 
      </ul> 
    <div style="clear:both";> 
<? 

var_dump($DadosProduto); 

if($DadosProduto["Imagem" . $_GET["I"]] != "") { 

    print '<div class="panel panel-default" style="width:145px; float:left; margin-right:4px;">'; 
    print ' <div class="panel-body">'; 
    print '  <img src="../gdThumb.php?imagem=' . $DadosProduto["Imagem" . $_GET["I"]] . '" alt="" class="img-thumbnail">'; 
    print ' </div>'; 
    print ' <div class="panel-footer"><center><a href="?acao=excluir&k='. $DadosProduto["idBanner"] .'&I='. $_GET["I"] .'"><span class="glyphicon glyphicon-remove"></span> Remover</a></center></div>'; 
    print '</div> '; 

} 


?> 
    </div> 
</form> 

     <script src="upload/assets/js/jquery.knob.js"></script> 

     <!-- jQuery File Upload Dependencies --> 
     <script src="upload/assets/js/jquery.ui.widget.js"></script> 
     <script src="upload/assets/js/jquery.iframe-transport.js"></script> 
     <script src="upload/assets/js/jquery.fileupload.js"></script> 

     <!-- Our main JS file --> 
     <script src="upload/assets/js/script.js"></script> 

<? require_once("footer.php"); ?> 

我认为这个问题是在表单动作

<form class="form-horizontal" role="form" id="upload" action="upload/uploadHome.php?k=<?=$id; ?>&I=<?=$_GET['I']; ?>" method="post" enctype="multipart/form-data"> 

因为上传文件夹中存在和PHP文件了。但是当表单调用这个动作时，什么都不会发生。我试图使用：var_dump，die（），echo，很多事情来看一个简单的反应。我无法调试uploadHome.php，我不知道为什么。

的connFile.php工作

所以，我需要调用此页面来存储我的形象和页面并没有叫。这里是uploadHome.php：

<? require_once("../ConnFile.php"); ?> 
<?php 
$idProduto = $_GET["k"]; 
// A list of permitted file extensions 
$allowed = array('png', 'jpg', 'gif','zip'); 

if(isset($_FILES['upl']) && $_FILES['upl']['error'] == 0){ 

    $extension = pathinfo($_FILES['upl']['name'], PATHINFO_EXTENSION); 

    if(!in_array(strtolower($extension), $allowed)){ 
     echo '{"status":"error"}'; 
     exit; 
    } 
    $arquivo = 'BannerHome_'. $_GET["k"] . '_' . rand(1,1500) . '-' . md5($_FILES['upl']['name']) . "." . $extension; 
    if(move_uploaded_file($_FILES['upl']['tmp_name'], '../../img/produto/'. $arquivo)){ 

     mysqli_query($link, "UPDATE BannerHome SET Imagem". $_GET['I'] ." = '$arquivo' WHERE idBanner = " . $_GET["k"]); 

     echo '{"status":"success"}'; 
     exit; 
    } 
} 

echo '{"status":"error"}'; 
exit; 

你能提供给我任何信息，将不胜感激，因为我不知道该怎么办。

Answer 1

我有权限错误。我看不到这个错误，因为阿贾克斯没有向我展示任何东西。

如果任何人有类似的问题，检查和更改目录权限可能会有所帮助。