填写一个表单,其中包含两部分,文件名和文本。我需要在html目录中创建一个名为$filename的文件,并获取我放入$text中的文本。不幸的是,我无法使它工作有什么帮助吗:
表单代码:
<form method="post" action="createfile.php">
<font size"3"><b>File name:</b> </font>
<br /><input type="text" size="15" name="filename" id="filename"><br />
<font size"3"><b>File data:</b> </font>
<div id="dat"><textarea name="thedata" id="thedata" cols="30" rows="15"></textarea></div>
<div id="sub" style="position:absolute; left: 25">
<input type="submit" value=" Create ">
</div>
</form>
createfile.php代码:
<?php
$name= $filename;
$data= $thedata;
if ( $file && $data ) {
$fp = fopen($file, "w");
fwrite($fp, "<br>");
fwrite($fp, $data);
fclose($fp);
}
else
{ echo 'no text entered';
}
?>
感谢的帮助
HTML将相同:
<form method="post" action="createfile.php">
<font size"3"><b>File name:</b> </font>
<br /><input type="text" size="15" name="filename" id="filename"><br />
<font size"3"><b>File data:</b> </font>
<div id="dat"><textarea name="thedata" id="thedata" cols="30" rows="15"></textarea></div>
<div id="sub" style="position:absolute; left: 25">
<input type="submit" value=" Create ">
</div>
</form>
PHP:
<?php
$filename = $_POST['filename'];
$content = $_POST['thedata'] ;
$if(!isset($_POST['filename'])){
//i leave the filename && content validation to the client side
$handle = fopen("/files/$filename.txt", "w");
fwrite($handle, $content);
fclose($handle);
}else{
//your `else` code here.
}
?>
使用以下代码:
<?php
$name= isset($_POST['filename'])?$_POST['filename']:0;
$data= isset($_POST['thedata'])?$_POST['thedata']:0;
if ($name && $data ) {
$fp = fopen($name, "w");
fwrite($fp, "<br>");
fwrite($fp, $data);
fclose($fp);
}
else {
echo 'no text entered';
}
?>