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我的表单中有文件上传和输入字段..如果文件上传成功，我想隐藏或禁用文件上传的提交按钮...我真的很困惑如何在同一页面上显示错误消息和如果没有错误，请转到“谢谢”页面。我正在使用 php 验证文件上传。如果文件上传成功，关于如何禁用提交按钮的任何想法？或向我展示如何同时处理文件上传和输入字段，同时在同一页面上显示错误消息以进行文件上传...

注意我希望文件上传的提交按钮只有在文件成功上传时才会消失，而不是在用户点击提交时消失。

<html>
<head>
</head>
<body>
<form action="" method="post" enctype="multipart/form-data">
<label for="file">Choose Photo:</label>
<input type="file" name="file" onchange="file_selected = true;">
<input type="hidden" name="submited" value="true" /> 
<input type="submit" name="submit" value="Submit" >
</form>


<form action="Send.php" method="post">
First Name:<input type="text" name="fname" required><br>
Last Name:<input type="text" name="lname" required><br>
Choose Username:<input type="text" name="username" required><br>
Age:<input type="text" name="age" required><br>
<input type="submit" value="Submit" name="submit">
</form>
</body>
</html>

这是处理文件上传的 php 代码...我有两个提交按钮，一个用于文件上传，一个用于其他输入字段..此 php 代码与 html 表单位于同一页面上。

<?php
ini_set( "display_errors", 0);
if(isset($_REQUEST['submited'])) {

// your save code goes here


$allowedExts = array("jpg", "jpeg", "gif", "png");
$extension = end(explode(".", $_FILES["file"]["name"]));
if ((($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/jpeg")
|| ($_FILES["file"]["type"] == "image/png")
|| ($_FILES["file"]["type"] == "image/pjpeg"))
&& ($_FILES["file"]["size"] < 2097152)
&& in_array($extension, $allowedExts))
{
if ($_FILES["file"]["error"] > 0)
{
echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
}
else
{
echo "";

if (file_exists("images/" . $_FILES["file"]["name"]))
{
echo "<font color='red'><b>We are sorry, the file you trying to upload already exists.</b></font>";
  }

else
{
move_uploaded_file($_FILES["file"]["tmp_name"],
"images/" . $_FILES["file"]["name"]);
echo "<font color='green'><b> Success! Your photo has been uploaded.</b></font>";
}
}
}
else
{
echo "<font color='red'><b>We are sorry, the file you trying to upload is not an image or it exceeds 2MB in size.</b></font><br><font color='blue'><i>Only images under size of 2MB are allowed</i></font>.";
}
}
?>
4

3 回答 3

1

我会使用 Jquery AJAX 调用将数据发送到 PHP 脚本。然后从响应中检索一个布尔值以确定按钮是否应该可见。

于 2013-01-01T04:07:49.293 回答
0

您可以通过简单的方式来做到这一点，例如。

    <html>
    <head>
    </head>
    <body>
    <?php
$sub=0;
    ini_set( "display_errors", 0);
    if(isset($_REQUEST['submited'])) {

    // your save code goes here

    $allowedExts = array("jpg", "jpeg", "gif", "png");
    $extension = end(explode(".", $_FILES["file"]["name"]));
    if ((($_FILES["file"]["type"] == "image/gif")
    || ($_FILES["file"]["type"] == "image/jpeg")
    || ($_FILES["file"]["type"] == "image/png")
    || ($_FILES["file"]["type"] == "image/pjpeg"))
    && ($_FILES["file"]["size"] < 2097152)
    && in_array($extension, $allowedExts))
    {
    if ($_FILES["file"]["error"] > 0)
    {
    echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
    }
    else
    {
    echo "";

    if (file_exists("images/" . $_FILES["file"]["name"]))
    {
    echo "<font color='red'><b>We are sorry, the file you trying to upload already exists.</b></font>";
      }

    else
    {
    move_uploaded_file($_FILES["file"]["tmp_name"],
    "images/" . $_FILES["file"]["name"]);
    $sub= 1;
    echo "<font color='green'><b> Success! Your photo has been uploaded.</b></font>";
    }
    }
    }
    else
    {
    echo "<font color='red'><b>We are sorry, the file you trying to upload is not an image or it exceeds 2MB in size.</b></font><br><font color='blue'><i>Only images under size of 2MB are allowed</i></font>.";
    }
    }
    ?>
    <form action="" method="post" enctype="multipart/form-data">
    <label for="file">Choose Photo:</label>
    <input type="file" name="file" onchange="file_selected = true;">
    <input type="hidden" name="submited" value="true" />

    <input type="submit" name="submit" value="Submit" >
    </form>


    <form action="Send.php" method="post">
    First Name:<input type="text" name="fname" required><br>
    Last Name:<input type="text" name="lname" required><br>
    Choose Username:<input type="text" name="username" required><br>
    Age:<input type="text" name="age" required><br>
<?php
if($sub==0)
{
?>
    <input type="submit" value="Submit" name="submit">
<?php
}
?>
    </form>
    </body>
    </html>

我假设您的代码是正确的。$sub=0我在开始时初始化了一个变量。如果succesfully uploaded设置为 1。

结束，如果$sub不等于零，则不显示提交。

所以，如果文件上传成功。按钮不会显示。

于 2013-01-01T04:09:31.197 回答
0 
<form action="**somepage.php**" method="post" enctype="multipart/form-data">
    <label for="file">Choose Photo:</label>
    <input type="file" name="file" onchange="file_selected = true;">
    <input type="hidden" name="submited" value="true" /> 
    <input type="submit" name="submit" value="Submit" >
    </form>

并从那里将指示不同状态的变量传递回主页，并根据变量显示错误消息。

如果您确实需要禁用input type file将变量保存在隐藏输入中并在页面加载时使用Jquery并禁用input type file....

没有jquery它的简单，只是在输入类型表单之前给出一个条件

    if($_GET['status']=successful)
    {
    <input type=file readonly="readonly" />
    }
   else
    {
    <input type=file />
    }
于 2013-01-01T04:58:48.683 回答