2014-02-18 13 views
-1

你好傢伙我需要這段代碼的幫助我想做一個用戶日誌,輸出爲.txt 我的問題是當我點擊按鈕。示例VIEW按鈕。下面的if語句不工作....不知道爲什麼..我該如何在if語句裏面運行我的代碼當我點擊按鈕

我嘗試把這個代碼

echo '<FORM METHOD="POST" ACTION="mydata2.php"> 
<INPUT TYPE="submit" name="submit" value="View"> 
</FORM>'; 

if(isset($_POST['submit']) && ($_POST['submit'] == "View")) 
{ 
$date=date("Y-m-d H:i:s"); 
$updatefile = "userlogs.txt"; 
$fh = fopen($updatefile, 'a') or die("can't open file"); 
$stringData = "User: $username"; 
fwrite($fh, "$stringData".PHP_EOL); 
$stringData = "Clicked View Button $date"; 
fwrite($fh, "$stringData".PHP_EOL); 
fclose($fh); 
} 

遺憾的英語不好..我要運行,當你點擊上面的代碼查看按鈕。沒有你們明白我的意思>; /我不能得到它..它不工作

+0

上面的PHP代碼是來自'mydata2.php'還是其他文件? – Mike

+0

什麼不起作用? if語句是否有效?它試圖寫入文件並失敗? – andrewsi

+0

首先想到的是:JavaScript執行客戶端,PHP服務器端。您可能想用'FILE_APPEND'標誌查看[file_put_contents](http://us2.php.net/function.file-put-contents)。 –

回答

0

試試這個

file1.php

echo '<from method="POST" action="mydata2.php"> 
<input type="submit" name="submit" value="View"> 
</form>'; 

mydata2.php

在這個$username沒有定義,你應該定義它。

if(isset($_POST['submit']) && ($_POST['submit'] == "View")) 
{ 
    $username = "your_username"; // define it 

    $date=date("Y-m-d H:i:s"); 
    $updatefile = "userlogs.txt"; 
    $fh = fopen($updatefile, 'a') or die("can't open file"); 
    $stringData = "User: $username"; 
    fwrite($fh, $stringData.PHP_EOL); 
    $stringData = "Clicked View Button $date"; 
    fwrite($fh, $stringData.PHP_EOL); 
    fclose($fh); 
} 
+0

是的!先生,謝謝你!那幫助我很多:))))) – Mrtata01

0

我試過這段代碼它正在工作。你可能會錯過你的數據。我有硬編碼的數據。

<?php 

echo '<FORM METHOD="POST" ACTION="mydata2.php"> 
<INPUT TYPE="submit" name="submit" value="View"> 
</FORM>'; 

if(isset($_POST['submit']) && ($_POST['submit'] == "View")) 
{ 

    $username = 'test'; 
    $date = date('Y-m-d'); 
    $stringData = 'String data'; 
    $date=date("Y-m-d H:i:s"); 
    $updatefile = "userlogs.txt"; 
    $fh = fopen($updatefile, 'a') or die("can't open file"); 
    $stringData = "User: $username"; 
    fwrite($fh, "$stringData".PHP_EOL); 
    $stringData = "Clicked View Button $date"; 
    fwrite($fh, "$stringData".PHP_EOL); 
    fclose($fh); 
} 
0

這是測試和工作正常..請參閱我所做的更改。

<?php 
    echo '<form method="POST" action=""> 
     <input type="submit" name="submit" value="View"> 
     </form>'; 

if(isset($_POST['submit']) && ($_POST['submit'] == "View")) 
{ 
$date=date("Y-m-d H:i:s"); 
$updatefile = "userlogs.txt"; 
$fh = fopen($updatefile, 'a') or die("can't open file"); 
$stringData = "User: "; 
fwrite($fh, "$stringData".PHP_EOL); 
$stringData = "Clicked View Button $date"; 
fwrite($fh, "$stringData".PHP_EOL); 
fclose($fh); 
} 
?> 
+0

謝謝你的答案先生:) – Mrtata01

相關問題