点击执行mysql DELETE查询

Question

我是 php 和 sql 领域的新玩家。 单击删除链接（或按钮）时，我试图从我的人员表中删除身份 谁能告诉我我做错了什么？

这是我的 php 代码：

<?php
    $db = new DB();
    $cg_id = $_SESSION['cg_id'];
    $cg_address_id = $_SESSION['cg_address_id'];
    $sql ="SELECT f_name, phone, c.id as idc
           FROM contacts as c 
           WHERE c.cg_id = '$cg_id'";

    $result = $db->mysqli->query($sql);
    if ($result->num_rows > 0) {
        // output data of each row
        while($row = mysqli_fetch_assoc($result)) {
            echo "<article class='contactArea'>";
            echo  "<a href='contacts2.php?del=".$row["idc"]."' class='deleteContact' name='remove' value='remove'>Remove</a></article>";    

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

            $idco = $_POST['idc'];

            $removeQuery = "DELETE FROM contacts as c WHERE id=".$idco." ";


            $resultt = mysql_query($removeQuery);
            if($resultt) {
            header('Location: '.$_SERVER['REQUEST_URI']);
            }
            echo "<script>window.location.reload(true);</script>";
        }
        }

    }else {
        echo "Please edit senior profile for monitoring!";
    }
?>

Answer 1

试试这个（显然用你的 mysql 服务器和数据库的详细信息替换“localhost”、“dbuser”、“dbpassword”和“database_name”）：

<?php
                    $db = new mysqli("localhost","dbuser","dbpassword","database_name");
                    $cg_id = $_SESSION['cg_id'];
                    $cg_address_id = $_SESSION['cg_address_id'];
                    // I've moved the deletion code to BEFORE the select query, otherwise the
                    // query will be shown including the to-be-deleted data and it is then deleted after it is displayed
                    if(isset($_GET["del"])){ // <--- this was $_POST["del"] which would have been unset
                        $idc = $_GET["del"];
                        if($db->query("DELETE FROM contacts WHERE id=$idc")){
                             echo "deleted";
                        } else { 
                            echo "fail";
                        }    
                    }  
                    $sql ="SELECT photo, f_name, phone, street, street_num, city, l_name, c.id as idc FROM contacts as c, address as a WHERE c.cg_id = '$cg_id' and a.id = c.address_id";
                    $result = $db->query($sql);
                    if ($result->num_rows > 0) {
                        // output data of each row
                        while($row = mysqli_fetch_assoc($result)) {
                            echo "<article class='contactArea'>";
                            echo "<article class='contact5 lior'>";
                            echo "<img class='CSImage' src='" .$row["photo"]."'>";
                            echo "<section class='generalFormTextW nameCPosition'> " .$row["f_name"]." ".$row["l_name"]."<br></section>";
                            echo "<section class='generalFormTextW phoneCPosition'> " .$row["phone"]."<br></section>";
                            echo "<section class='generalFormTextB addressCPosition'>".$row["city"].", <br> ".$row["street"]." ".$row["street_num"]. "<br></section>";
                            echo  "<a href='contacts2.php?del=".$row["idc"]."' class='deleteContact' name='remove' value='remove'>Remove</a></article></article>";       
                        }
                    }
?>

请注意，我正在更改您使用 mysqli 的方式，以便您直接使用它，而不是作为 DB 对象的成员，这是我在其他地方看到的使用方式 - 在我看来，好像您实际上并没有打开数据库连接（尽管您可能只是没有包含它，因为它显示了您的密码？）

**编辑：我已将 $_POST["del"] 更改为 $_GET["del"] - 因为您在 url ("contacts2.php?del=") 中设置 del 这将是 GET不是 POST。

**编辑：我已经移动了删除代码，以便它解决您必须刷新页面才能查看已删除记录的数据的问题 - 以前显示信息然后删除，我们要删除 THEN显示。