PHP 中 foreach 循环的每次迭代的提交按钮

Question

我对编码很陌生，所以请轻轻地...

我正在创建一个管理页面，其中包含需要验证的帐户列表。 我只需要在单击按钮时将 DB 中“已验证”列中的值从 0 更改为 1。 问题是它会为循环中的所有其他返回结果触发相同的事情，每个人都有自己的按钮，而不仅仅是循环的特定迭代。 任何帮助将不胜感激。 代码目前如下所示：

<?php
$sql ="SELECT customer.First_Name, customer.Last_Name, account.account_no, account.client_id, customer.username
FROM customer 
 INNER JOIN account 
 ON customer.customer_id=account.client_id 
 WHERE validated = 0"; 

$tobe_validated = $dbh->query($sql);

foreach ($tobe_validated as $row) {
        //creating variable for account number to put in query  
        $clientid=$row["client_id"];

        echo "<div class='valid_name_btn'>";  
        echo "<form method='post'><input type='submit' class='btn btn-outline-primary' value ='Validate' name='validate' id='validate'></input></form>";
        echo "<div class='valid_name'>"; //div for name
        echo $row["First_Name"] . " " . $row["Last_Name"]." - Account No. ". $row["account_no"]."<br/>"; //show name and account number of client
        echo "</div>";
        echo "</div>";


        // query to change validated in customer table to 1
          $sql ="UPDATE customer SET validated = 1 WHERE customer_id = '$clientid'";

        // validate account when button is clicked
          if(isset($_POST['validate'])) {
          $dbh->query($sql);
          } 
         }

Answer 1

您有逻辑错误，因为您的表单只是传递validate值，而不是确切的clientid 。 您必须从循环中移出$_POST操作，并在每个表单中添加一个具有相应clientid的隐藏字段：

<?php
    $sql = "SELECT customer.First_Name, customer.Last_Name, account.account_no, account.client_id, customer.username
            FROM customer
            INNER JOIN account ON customer.customer_id=account.client_id
             WHERE validated = 0";

    $tobe_validated = $dbh->query($sql);

    foreach ($tobe_validated as $row) {
        //creating variable for account number to put in query
        $clientid = $row["client_id"];

        echo "<div class='valid_name_btn'>";
            echo "<form method='post'>
                <input type='hidden' name='clientid' value='".$clientid."'>
                <input type='submit' class='btn btn-outline-primary' value ='Validate' name='validate' id='validate'></input>
            </form>";
            echo "<div class='valid_name'>"; //div for name
                echo $row["First_Name"] . " " . $row["Last_Name"] . " - Account No. " . $row["account_no"] . "<br/>"; //show name and account number of client
            echo "</div>";
        echo "</div>";
    }

    if (isset($_POST['validate']) && isset($_POST['clientid'])) {
        $clientid = $_POST['clientid'];
        $sql = "UPDATE customer SET validated = 1 WHERE customer_id = '$clientid'";
        $dbh->query($sql);
    }

Answer 2

您可以根据键制作动态表单，并可以通过制作如下适当的条件来检查按钮是否具有特定值

<?php
$sql ="SELECT customer.First_Name, customer.Last_Name, account.account_no, account.client_id, customer.username
FROM customer 
 INNER JOIN account 
 ON customer.customer_id=account.client_id 
 WHERE validated = 0"; 

$tobe_validated = $dbh->query($sql);

foreach ($tobe_validated as $key => $row) {
        //creating variable for account number to put in query  
        $clientid=$row["client_id"];

        echo "<div class='valid_name_btn'>";  
        echo "<form method='post' name="'validation_form_'.$key"><input type='submit' class='btn btn-outline-primary' value ='Validate' name="'validate_'.$key" id="'validate_'.$key"></input></form>";
        echo "<div class='valid_name'>"; //div for name
        echo $row["First_Name"] . " " . $row["Last_Name"]." - Account No. ". $row["account_no"]."<br/>"; //show name and account number of client
        echo "</div>";
        echo "</div>";

        // validate account when button is clicked
          if(isset($_POST['validate_'.$key]) && $_POST['validate_'.$key] == 'Validate') {
            // query to change validated in customer table to 1
            $updateSql ="UPDATE customer SET validated = 1 WHERE customer_id = '$clientid'";
            $dbh->query($updateSql);
          } 
         }

如果您需要任何帮助，请告诉我

Answer 3

首先，您的表单 - 提交时 - 没有提供有关要验证的客户 ID 的任何线索。 在我的示例代码中，我不会使用 echo 因为像这样回显 HTML 输出是丑陋的。 相反，您可以这样做：

?>
<div class='valid_name_btn'>
  <form method='post'>
    <input type="hidden" name="client" value="<?= $clientid ?>">
    <input type='submit' class='btn btn-outline-primary' value ='Validate' name='validate' id='validate'></input>
  </form>
  <div class='valid_name'>
    <?= $row["First_Name"] . " " . $row["Last_Name"]." - Account No. ".$row["account_no"] ?><br/>
  </div>
</div>
<?php

我在表单中添加了一个隐藏的输入字段，然后您可以在文件的开头，在$tobe_validated = ..行之前对其进行评估：

if (isset($_POST['validate']) && (isset($_POST['client'])) {
  $clientid = $_POST['client'];
  $sql = "UPDATE customer SET validated = 1 WHERE customer_id = '$clientid'";
  $dbh->query($sql);
}

您还应该正确准备语句，或者至少转义 $client 值。 因为我不知道 $dbh 是什么（你的代码没有告诉）我不会在这里做这个。

Answer 4

$sql ="UPDATE customer SET validated = 1 WHERE customer_id = $clientid";

$clientid应该没有 ''