Question

我只是想将一些表单数据输出到文本文件中。我无法让它发挥作用。有人能帮忙吗？我收到错误invalid offset type。谢谢你的帮助！

这是php

$file = "output.txt";

$current = file_get_contents($file);

foreach ($_POST as $key) {
    $current .= $_POST["data"][$key];
}

file_put_contents($file, $current);

这是html

<form name="input" action="getinput.php" method="post">
  Name: <br /><input type="text" name="data[]" placeholder="First and Last"><br />
  ID: <br /><input type="text" name="data[]" placeholder="Your ID"><br />
  Major: <br /><input type="text" name="data[]" placeholder="Your Major"><br />
  Email: <br /><input type="email" name="data[]" placeholder="Email Address"><br />
  Phone: <br /><input type="text" name="data[]" placeholder="XXX-XXX-XXXX"><br />
  Addition Information: <br /><textarea name="data[]" cols="30" rows="10" placeholder="Please enter any extra  or comments you would like me to have or know."></textarea><br />
  <input type="submit" value="Submit">
</form>

Answer 1

怎么样：

foreach ($_POST['data'] as $key => $value) {
   $current .= $value;
}

现在您循环遍历整个$_POST数组，而不仅仅是包含$_POST['data']的部分，这就是您在执行此操作时收到Invalid offset警告的原因：

$_POST["data"][$key];

Answer 2

您的foreach不对。你想要这样的东西输出值而不是键

foreach($_POST['data'] as $value) {

    $current .= $value;
}

在PHP中将表单数据输出到文件

2 个答案: