PHP发送表单数据
我新的PHP和形式的发展,这里就是我想实现:PHP发送表单数据
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:)
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>" . $fileName . "\n";
}
?>
<input type="submit" name="finalSubmit" value="Submit">
</form>
他们不需要,当你填写在浏览器中,他们将被添加 – EaterOfCode 2012-08-06 12:11:48
THX的答案 – Arek 2012-08-06 12:15:11
预填充它们对,认为它必须以其他方式填充,而不是手动,然后忽略价值属性部分。别客气。 – 2012-08-06 12:16:37
替换您gather.php与
<br>
Final form
<br>
<form action="build.php" method="post">
<?php
if(isset($_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>" . $fileName . "\n";
}
?>
<input type="submit" name="finalSubmit" value="Submit">
</form>
你是echo'ing形式之外的输入框因此现在将工作
大声笑,谢谢你的答案:) – Arek 2012-08-06 12:09:53
欢迎:) – EaterOfCode 2012-08-06 12:12:19
我觉得<form>
第二种形式的需要来到文件的顶部 - 它只会在标签内提交元素,因为您正在生成HTML并打开表单,所以未提交。
<br>
Final form
<br>
<form action="build.php" method="post">
<?php
if(isset($_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>
thx为答案 – Arek 2012-08-06 12:16:56
您需要再次添加表单元素,你的第二个形式,并用正确的价值观 – 2012-08-06 12:03:52