如何將所有參數添加到表單中的.txt？

Question

我的問題是，我想將下面顯示的所有值寫入一個txt文件。

我收到一個錯誤，當這樣做是在線fwrite(...) - 它只能除了3個參數。

我做錯了什麼？ 有沒有更好的方法來做到這一點？

謝謝你的幫助。

<?php 
if(isset($_POST['submit'])) 
{ 
    $name = $_POST['name']; 
    $gender = $_POST['gender']; 
    $age = $_POST['age']; 
    $personality = $_POST['personality']; 
    $OS = $_POST['OS']; 
    $age_min = $_POST['age_min']; 
    $age_max = $_POST['age_max']; 

    $file = fopen("singles.txt","a+"); 
    fwrite($file,$name,$gender,$age,$personality,$OS,$age_min,$age_max); 
    fclose($file); 
    print_r(error_get_last()); 
} 
?> 

如果需要其他信息，我可以嘗試幫忙！

Answer 1

函數fwrite期望您傳遞文件指針（在您的情況下爲$ file）和要寫入的字符串。如果你想有一個逗號分隔的列表，那麼你應該做這樣的：

fwrite($file,$name . "," . $gender . "," . $age . "," . $personality . "," . $OS . "," . $age_min . "," . $ag);

http://php.net/manual/en/function.fwrite.php

http://php.net/manual/en/language.operators.string.php

Answer 2

<?php 
if(isset($_POST['submit'])) 
{ 
    $name = $_POST['name']; 
    $gender = $_POST['gender']; 
    $age = $_POST['age']; 
    $personality = $_POST['personality']; 
    $OS = $_POST['OS']; 
    $age_min = $_POST['age_min']; 
    $age_max = $_POST['age_max']; 

    $variable_str = $name.",".$gender.",".$age.",".$personality.",".$os.",".$age_min.",".$age_max; 

    $file = fopen("singles.txt","a+"); 
    fwrite($file,$variable_str); 
    fclose($file); 
    print_r(error_get_last()); 
} 
?> 

你應該把這些值的字符串，然後通過轉換成fwrite（）函數。關於fwrite（）Link的更多細節就在這裏。