PHP发送表单数据

Question

我新的PHP和形式的发展，这里就是我想实现：

首先我有一个简单的表格，输入刚才两个文本值：

Form1 
<br> 
<form action="gather.php" method="post"> 
    Catalog: 
    <input type="text" name="folderName" maxlength="50"> 
    <br> 
    File Name: 
    <input type="text" name="fileName" maxlength="50"> 
    <br> 
    <input type="submit" name="formSubmit" value="Submit"> 
</form> 

现在我有一个名为gather.php第二个文件从哪里获得theese两条线，并利用它们来算目录里面等

<?php 
if(isset($_POST['formSubmit'])){ 
    $folderName = $_POST['folderName']; 
    $fileName = $_POST['fileName']; 
    $numberOfImages = count(glob($folderName . "/*.jpg")); 
    for($i = 1; $i <= $numberOfImages; $i++){ 
     echo "<br><input type=\"text\" name=\"imie" . $i . "\"><br/>\n"; 
     echo "<img src=\"" . $folderName . "/0" . $i . ".jpg\" height=\"50px\" width=\"50px\"><br><br>\n"; 
    } 


    echo "\n<br>" . $folderName . "<br>" . $fileName . "\n"; 
} 

?> 
<br> 
Final form 
<br> 
<form action="build.php" method="post"> 
<input type="submit" name="finalSubmit" value="Submit"> 
</form> 

文件，这应该让我build.php文件，它看起来更不像这样的：

<?php 
if(isset($_POST['finalSubmit'])){ 
    //loop and other stuff 
    $temp = $_POST['imie1']; 
    echo $temp; 

} 
?> 

所以事情是，在这最后的文件我想获得所有被放入文本字​​段中gather.php文件中的数据。但是我得到了build.php上未定义的索引错误，说$ _POST ['imie1']中没有任何内容。你能告诉我为什么吗？这是从第二个文件到第三个文件的数据吗？

编辑：THX的答案，因为我只能接受1，多是我选择具有至少代表用户只是为了支持同她:)

Answer 1

您需要添加表单标签内的输入，它不会被发送，否则。

<br> 
    Final form 
    <br> 
    <form action="build.php" method="post"> 
    <?php 
    if(isset($_POST['formSubmit'])){ 
     $folderName = $_POST['folderName']; 
     $fileName = $_POST['fileName']; 
     $numberOfImages = count(glob($folderName . "/*.jpg")); 
     for($i = 1; $i <= $numberOfImages; $i++){ 
      echo "<br><input type=\"text\" name=\"imie" . $i . "\"><br/>\n"; 
      echo "<img src=\"" . $folderName . "/0" . $i . ".jpg\" height=\"50px\" width=\"50px\"><br><br>\n"; 
     } 


     echo "\n<br>" . $folderName . "<br>" . $fileName . "\n"; 
    } 

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

Answer 2

替换您gather.php与

<br> 
Final form 
<br> 
<form action="build.php" method="post"> 
<?php 
    if(isset($_POST['formSubmit'])){ 
     $folderName = $_POST['folderName']; 
     $fileName = $_POST['fileName']; 
     $numberOfImages = count(glob($folderName . "/*.jpg")); 
     for($i = 1; $i <= $numberOfImages; $i++){ 
      echo "<br><input type=\"text\" name=\"imie" . $i . "\"><br/>\n"; 
      echo "<img src=\"" . $folderName . "/0" . $i . ".jpg\" height=\"50px\" width=\"50px\"><br><br>\n"; 
     } 


     echo "\n<br>" . $folderName . "<br>" . $fileName . "\n"; 
    } 

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

你是echo'ing形式之外的输入框因此现在将工作

Answer 3

我觉得<form>第二种形式的需要来到文件的顶部 - 它只会在标签内提交元素，因为您正在生成HTML并打开表单，所以未提交。

<br> 
Final form 
<br> 
<form action="build.php" method="post"> 
<?php 
if(isset($_POST['formSubmit'])){ 
    $folderName = $_POST['folderName']; 
    $fileName = $_POST['fileName']; 
    $numberOfImages = count(glob($folderName . "/*.jpg")); 
    for($i = 1; $i <= $numberOfImages; $i++){ 
     echo "<br><input type=\"text\" name=\"imie" . $i . "\"><br/>\n"; 
     echo "<img src=\"" . $folderName . "/0" . $i . ".jpg\" height=\"50px\" width=\"50px\"><br><br>\n"; 
    } 


    echo "\n<br>" . $folderName . "<br>" . $fileName . "\n"; 
} 

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