Question

Isset（$ _ POST ['submit）无法正常工作，但是当我添加not操作（！）时，它工作正常。我不在哪里找到错误。这是我的代码

$get = isset($_GET['id']) ? $_GET['id'] : "";
$query = "SELECT * FROM `crew_info` WHERE `id` = '$get'";
$query_run = mysqli_query($conn,$query);

    if(!isset($_POST['submit'])) {

        $query_update = "UPDATE `crew_info` SET `crew_status` = 'NEW' WHERE `id` = '$get'";
        if ($query_update_run = mysqli_query($conn,$query_update)) {

            echo 'Crew Accepted';
        }

    }

这是提交表单

<form action="review.php" method="POST">

    <table border="2" align="center" width="600" cellspacing="3" cellpadding="4">
        <tr>
            <th>First Name</th>
            <th>Middle Name</th>
            <th>Last Name</th>
            <th>Date Added</th>
            <th>Status</th>
        </tr>
        <tr>';
        while($record = mysqli_fetch_assoc($query_run)) {
        echo "<tr>";
        echo "<th>".$record['first_name']."</th>";
        echo "<th>".$record['middle_name']."</th>";
        echo "<th>".$record['last_name']."</th>";
        echo "<th>".$record['date_added']."</th>";
        echo "<th>".$record['crew_status']."</th>";
        echo '</tr>';
      }
     echo '</table>';
     echo '<br><center><input type="submit" name="submit"></center>
</form>

Answer 1

您没有提交任何值。

<input type="submit" name="submit">

尝试这样的事情：

<input type="submit" name="submit" value="ok">

这段PHP检查表单元素的名称是＆＃34;提交＆＃34;有任何价值：

"if isset($_POST['submit'])"

但是，由于您没有为＆＃34;提交＆＃34;设置任何值，因此无法验证。但是，它会在您设置前面的感叹号时进行验证：

!if isset($_POST['submit'])"

这是因为你说，＆＃34;如果＆＃39;提交＆＃39;没有设置值，请执行以下操作＆＃34;。

希望这是有道理的。 :)

更新：您还需要修复其他几个错误才能使整个过程正常运行。这应该解决大部分问题。将它与您当前的代码进行比较：

<?php $get = isset($_GET['id']) ? $_GET['id'] : "";
$query = "SELECT * FROM crew_info WHERE id = '$get'";
$query_run = mysqli_query($conn,$query);

if(isset($_POST['submit'])) {

    $query_update = "UPDATE crew_info SET crew_status = 'NEW' WHERE id = '$get'";
    if ($query_update_run = mysqli_query($conn,$query_update)) {

        echo 'Crew Accepted';
    }

} ?>

<form action="review.php" method="POST">

<table border="2" align="center" width="600" cellspacing="3" cellpadding="4">
    <tr>
        <th>First Name</th>
        <th>Middle Name</th>
        <th>Last Name</th>
        <th>Date Added</th>
        <th>Status</th>
    </tr>
    <tr>

<?php
    while($record = mysqli_fetch_assoc($query_run)) {
    echo "<tr>";
    echo "<th>".$record['first_name']."</th>";
    echo "<th>".$record['middle_name']."</th>";
    echo "<th>".$record['last_name']."</th>";
    echo "<th>".$record['date_added']."</th>";
    echo "<th>".$record['crew_status']."</th>";
    echo '</tr>';
  }
echo '</table>';
echo '<br><center><input type="submit" name="submit" value="ok"></center>';
?>

</form>

Isset（$ _ POST ['submit）不起作用

1 个答案: