Question

我在stackOverflow上发现了几个线程，这些线程解决了如何向PHP结果表添加“删除”按钮的问题，但是似乎没有一个线程可以解决我的特定用例。

我有一个从MySql数据库中提取的PHP结果表。我已经向表的每一行添加了“删除”链接，但是当我单击给定行上的“删除”链接时，它将删除表中的所有行。

我怀疑我没有正确实现'delete'链接，但是我对PHP不够熟练，无法弄清楚到底是什么问题。

下面是生成表的代码。请注意，在endChildren（）函数中回显了“删除”链接。

type GetKeys<T extends readonly {key: any}[]> = T[number]['key'];

type Keys = GetKeys<typeof arr>;
// type Keys = "one" | "two" | "three"

这是delete.php脚本：

        <?php
        echo "<table style='border: solid 1px black;'>";
        echo "<tr><th>Id</th><th>Name</th><th>Number</th><th>Part A</th><th>Part B</th><th>Full Name</th><th>Address</th><th>Apt.</th><th>City</th><th>State</th><th>Zip</th><th>Remove</th></tr>";

        class TableRows extends RecursiveIteratorIterator { 
            function __construct($it) { 
                parent::__construct($it, self::LEAVES_ONLY); 
            }

            function current() {
                return "<td>" . parent::current(). "</td>";
            }

            function beginChildren() { 
                echo "<tr>"; 
            } 

            function endChildren() { 
                echo "<td><a href='delete.php?id=".$row['id']."'>Delete></a></td>";
                echo "</tr>" . "\n";
            } 
        } 


        $servername = "localhost:3306";
        $username = "xxxxxxxxxx";
        $password = "xxxxxxxx";
        $dbname = "xxxxxxxxx";

        try {
            $conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
            $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
            $stmt = $conn->prepare("SELECT id, Name, Number, PartA, PartB, Full_Name, Address, Apt, City, State, Zip FROM Medicard"); 
            $stmt->execute();

            // set the resulting array to associative
            $result = $stmt->setFetchMode(PDO::FETCH_ASSOC); 

            foreach(new TableRows(new RecursiveArrayIterator($stmt->fetchAll())) as $k=>$v) { 
                echo $v;
            }

        }
        catch(PDOException $e) {
            echo "Error: " . $e->getMessage();
        }
        $conn = null;
        echo "</table>";
        ?>

当我单击给定行中的删除链接时，它将删除表中的所有行。我不确定这是因为我在其中添加了“删除”链接，还是我的delete.php代码不正确。

我对PHP和MySql比较陌生。任何指导将不胜感激。谢谢。

Answer 1

您删除列id与列id匹配的所有行-当然，所有行都是这种情况。

$sql = "DELETE FROM Medicard WHERE id=id";

您需要从GET参数中读取ID并调整您的查询：

  <?php
    $servername = "localhost:3306";
    $username = "xxxxxxxx";
    $password = "xxxxxxxx";
    $dbname = "xxxxxxxx";

    $id = $_GET["id"];

    try {
        $conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
        // set the PDO error mode to exception
        $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

        // sql to delete a record
        $sql = $conn->prepare("DELETE FROM Medicard WHERE id = ?");
        $success = $sql->execute(array($id));

        echo $success ? "Record deleted successfully" : "Record not deleted";
        }
    catch(PDOException $e)
        {
        echo $e->getMessage();
        }

    $conn = null;
    ?>

表生成部分（我用一个简单的循环替换了您实现的迭代器）

<?php
        echo "<table style='border: solid 1px black;'>";
        echo "<tr><th>Id</th><th>Name</th><th>Number</th><th>Part A</th><th>Part B</th><th>Full Name</th><th>Address</th><th>Apt.</th><th>City</th><th>State</th><th>Zip</th><th>Remove</th></tr>";

        $servername = "localhost:3306";
        $username = "xxxxxxxxxx";
        $password = "xxxxxxxx";
        $dbname = "xxxxxxxxx";

        try {
            $conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
            $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
            $stmt = $conn->prepare("SELECT id, Name, Number, PartA, PartB, Full_Name, Address, Apt, City, State, Zip FROM Medicard"); 
            $stmt->execute();

            // set the resulting array to associative
            $result = $stmt->setFetchMode(PDO::FETCH_ASSOC);

            foreach ($stmt->fetchAll() as $row) {
                echo "<tr>"; 
                foreach ($row as $val) {
                    echo "<td>" . $val . "</td>";
                }
                echo "<td><a href='delete.php?id=".$row['id']."'>Delete></a></td>";
                echo "</tr>";
            } 

        }
        catch(PDOException $e) {
            echo "Error: " . $e->getMessage();
        }
        $conn = null;
        echo "</table>";
?>

如何将删除按钮添加到php结果表的每一行

1 个答案: