PHP發送表單數據

Question

我新的PHP和形式的發展，這裏就是我想實現：

首先我有一個簡單的表格，輸入剛纔兩個文本值：

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> 

現在我有一個名爲gather.php第二個文件從哪裏獲得theese兩條線，並利用它們來算目錄裏面等

<?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，這應該讓我build.php文件，它看起來更不像這樣的：

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

} 
?> 

所以事情是，在這最後的文件我想獲得所有被放入文本字​​段中gather.php文件中的數據。但是我得到了build.php上未定義的索引錯誤，說$ _POST ['imie1']中沒有任何內容。你能告訴我爲什麼嗎？這是從第二個文件到第三個文件的數據嗎？

編輯：THX的答案，因爲我只能接受1，多是我選擇具有至少代表用戶只是爲了支持同她:)

Answer 1

您需要添加表單標籤內的輸入，它不會被髮送，否則。

<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

替換您gather.php與

<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形式之外的輸入框因此現在將工作

Answer 3

我覺得<form>第二種形式的需要來到文件的頂部 - 它只會在標籤內提交元素，因爲您正在生成HTML並打開表單，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>