如何使用 php 编辑/更新 txt 文件
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How to edit/update a txt file with php
提问by Mr Andreev
After I've read a lot of similar problems with the edit/update function on a file and none of it worked I would like to ask for some help.
在我阅读了许多关于文件的编辑/更新功能的类似问题并且没有一个工作后，我想寻求一些帮助。
I am trying to edit a .txtdocument from php. I have tried these things:
我正在尝试.txt从 php编辑文档。我试过这些东西：
This was the last code which I've read here and didn't work.
$data_to_write = "$_POST[subject]"; $file_path = "text/" + $row['name']; $file_handle = fopen($file_path, 'w'); fwrite($file_handle, $data_to_write); fclose($file_handle);And this is my previous try:
$new_contents = "$_POST[subject]\n"; $path = "text/$row[name]"; file_put_contents($path, $new_contents);
这是我在这里阅读的最后一个代码，但没有用。
$data_to_write = "$_POST[subject]"; $file_path = "text/" + $row['name']; $file_handle = fopen($file_path, 'w'); fwrite($file_handle, $data_to_write); fclose($file_handle);这是我之前的尝试：
$new_contents = "$_POST[subject]\n"; $path = "text/$row[name]"; file_put_contents($path, $new_contents);
I hope someone would explain me how to do this the right way. Thank you.
我希望有人能解释我如何以正确的方式做到这一点。谢谢你。
This is all of my code:
这是我所有的代码：
<?php
if(isset($_GET['id']))
{
$edit = mysql_query("SELECT * FROM text_area WHERE text_id=$_GET[id]");
$row = mysql_fetch_assoc($edit);
$contents = file_get_contents($row['content']);
?>
<form action="" name="form" method="post">
<input type="hidden" name="id" value="<?php echo $row['text_id']; ?>" /><br />
<label for="">Заглавие:</label><br />
<input type="text" name="title" style="width:500px;" value="<?php echo $row['subject'] ?>" /><br />
<select name="opt">
<option value="0"></option>
<?php
$result = mysql_query("SELECT * FROM image_area");
while ($row = mysql_fetch_array($result))
{
echo "<option value=" . $row['path'] . ">" . $row['name'] . "</option>
";
}
?>
</select><input type="button" name="sumbitP" value="Choose" onclick="addtext();" /><a href="../image_list.php" target="_blank">Image list</a><br />
<textarea rows="10" cols="50" name="text" id="markItUp"><?php echo $contents ?></textarea><br />
<input type="submit" name="sumbitT" value="Update" />
<input type="reset" value="Reset" />
</form>
<?php
}
?>
<?php
if(isset($_POST['id']))
{
if(mysql_query("UPDATE text_area SET title='$_POST[subject]' WHERE text_id ='$_POST[id]'"))
{
$data_to_write = "" . $_POST['text'];
$file_path = "text/$row[name]";
$file_handle = fopen($file_path, 'w');
fwrite($file_handle, $data_to_write);
fclose($file_handle);
echo '<br><br><p align="center">Everything is ok</p>';
} else {
echo '<br><br><p align="center">Everything is not ok</p>' ;
}
?>
Just to add something which might be useful:
I am getting this error which I can't manage to find an answer for with Google. Warning: fopen(text/) [function.fopen]: failed to open stream: Is a directory in
只是添加一些可能有用的东西：
我收到了这个错误，我无法通过 Google 找到答案。Warning: fopen(text/) [function.fopen]: failed to open stream: Is a directory in
回答by Nikhil Patel
You just need to change :
你只需要改变：
$file_path = "text/" + $row['name'];
to this :
对此：
$file_path = "text/" . $row['name'];
The concatenation operator in PHP is .(not +)
PHP 中的连接运算符是.(not +)
And make sure the directory textexists, otherwise its better to check and then write :
并确保目录text存在，否则最好检查然后写入：
$data_to_write = $_POST['subject'];
$file_path = "text/" . $row['name'];
if ( !file_exists("text") )
mkdir("text");
$file_handle = fopen($file_path, 'w');
fwrite($file_handle, $data_to_write);
fclose($file_handle);
回答by Ankur Kumar Singh
You can also open file in append mode using fopen() and put whatever you have at the end like
您还可以使用 fopen() 以追加模式打开文件，并将您拥有的任何内容放在最后，例如
$path = dirname(__FILE__).'/newfile.txt';
$fp = fopen($path, 'a');
if(!$fp){
echo 'file is not opend';
}
fwrite($fp, 'this is simple text written');
fclose($fp);
Detail is mentioned in TechFlirt File Handling Bloghere
此处的TechFlirt 文件处理博客中提到了详细信息
回答by Moeed Farooqui
You need to use file_get_contentsto get the text of your file.
您需要使用file_get_contents来获取文件的文本。
$file_path= "text/" . $row['name'];
// Open the file to get existing content
$current = file_get_contents($file_path);
// Append a new person to the file
$data_to_write.= $_POST[subject]."\n";
// Write the contents back to the file
file_put_contents($file_path, $data_to_write);
See Documentation
查看文档
回答by Moeed Farooqui
I just put the code into a compiler, you are missing a }at the end.
我只是将代码放入编译器中，最后您缺少一个}。
回答by user2781686
I think the problem may be the first line :
我认为问题可能出在第一行：
$data_to_write = "$_POST[subject]";
Replace it with the following :
将其替换为以下内容：
$data_to_write = "" . $_POST['subject'];
And i highly recommand to protect this with hmtlentities or anything else as it's a direct user input.
我强烈建议使用 hmtlentities 或其他任何东西来保护它，因为它是直接的用户输入。
