Question

以下代码可以让您查看朋友列表并在同一页面上编辑或删除。查看工作，但删除和更新没有。任何帮助，将不胜感激。查看代码：

$query ="SELECT * FROM tblFriends";
        $result = mysqli_query($conn,$query);
        while($row =$result->fetch_assoc()){

        $fname=$row['fname'];
        $lname=$row['lname'];
        $address=$row['address'];
        $desc=$row['description'];
        $zip=$row['zip'];
        $city=$row['city'];
        $state=$row['state'];
        $id =$row['key'];

    echo 
    "<tr>
    <td><input type='text' name='fname' value='$fname'/></td>
    <td><input type='text' name='lname' value='$lname'/></td>
    <td><input type='text' name='address' value='$address'/></td>
    <td><input type='text' name='city' value='$city'/></td>
    <td><input type='text' name='desc' value='$desc'/></td>
    <td><input type='text' name='state' value='$state'/></td>
    <td><input type='submit' name='Edit' value='Edit'/></td>
    <td><input type='submit' name='Delete' value='Delete'/></td>
    <input type='hidden' name='id' value='$id'/>
    <input type='hidden' name='zip' value='$zip'/>
    </tr>";
    }

更新代码：

        if(isset($_POST['Edit'])){
            $fname1=$_POST['fname'];
            $lname1=$_POST['lname'];
            $city1=$_POST['city'];
            $state1=$_POST['state'];
            $zip1=$_POST['zip'];
            $desc1=$_POST['desc'];
            $address1=$_POST['address'];
            $id1=$_POST['id'];
            $UpdateQuery ="UPDATE tblfriends SET fname='$fname1', lname='$lname1',city='$city1',address='$address1',zip='$zip1',state='$state1' WHERE id=$id1";
            echo $id1;
            if( mysqli_query($conn,$UpdateQuery)){
                echo "Updated";
            }else{
                echo "Not Updated";
            }

    }

删除代码：此mysqli_query和更新版本都返回false

if(isset($_POST['Delete'])){
            $id2 =$_POST['id'];
            $deleteQuery= "DELETE FROM tblfriends WHERE id=$id2";
            if( mysqli_query($conn,$deleteQuery)){
                echo "Deleted";
            }else{
                echo "Not Deleted";
            }

Answer 1

更改此查询"DELETE FROM tblfriends WHERE id=$id2" with "DELETE FROM tblfriends WHERE id='$id2'"。如果通过在查询编辑器中运行它来检查查询（如tab sql中的phpmyadmin），那就更好了。在代码中尝试这样的测试：

$deleteQuery= "DELETE FROM tblfriends WHERE id=$id2";
die($deleteQuery);

然后复制浏览器中显示的结果，然后粘贴到phpmyadmin中。在那里跑。希望它可以帮助你

Answer 2

我看了你的问题大声笑，把奇怪的人拿出去了？

$id =$row['key'];

"UPDATE tblfriends SET fname='$fname1', lname='$lname1',city='$city1',address='$address1',zip='$zip1',state='$state1' WHERE id=$id1

"DELETE FROM tblfriends WHERE id=$id2";

只是为了帮助你，你的身份名称是关键？但你检查id是否匹配id？不确定，但这可能是你的问题

因此请将您的查询更改为此

$query ="SELECT * FROM tblFriends";
$result = mysqli_query($conn,$query);
while($row = $result->fetch_assoc()){

    $fname=$row['fname'];
    $lname=$row['lname'];
    $address=$row['address'];
    $desc=$row['description'];
    $zip=$row['zip'];
    $city=$row['city'];
    $state=$row['state'];
    $id =$row['key'];

/**
**Wrapped your input into a form which will post the request
**/
    echo 
    "
    <form action=\"#\" method=\"POST\">
    <tr>
    <td><input type='text' name='fname' value='$fname'/></td>
    <td><input type='text' name='lname' value='$lname'/></td>
    <td><input type='text' name='address' value='$address'/></td>
    <td><input type='text' name='city' value='$city'/></td>
    <td><input type='text' name='desc' value='$desc'/></td>
    <td><input type='text' name='state' value='$state'/></td>
    <td><input type='submit' name='Edit' value='Edit'/></td>
    <td><input type='submit' name='Delete' value='Delete'/></td>
    <input type='hidden' name='id' value='$id'/>
    <input type='hidden' name='zip' value='$zip'/>
    </tr>
    <br>
    </form>
    ";
};

if(isset($_POST['Edit'])){
    $fname1= $_POST['fname'];
    $lname1= $_POST['lname'];
    $city1= $_POST['city'];
    $state1= $_POST['state'];
    $zip1= $_POST['zip'];
    $desc1= $_POST['desc'];
    $address1= $_POST['address'];
    $id1= $_POST['id'];



/**
**Updated your query to match the key rather than ID. `key`
**/

    $UpdateQuery ="UPDATE tblfriends SET fname='$fname1', lname='$lname1',city='$city1',address='$address1',zip='$zip1',state='$state1' WHERE `key` = '$id1' ";
    echo $id1;
    if( mysqli_query($conn,$UpdateQuery)){
        echo "Updated";
    }else{
        echo "Not Updated";
    };

};

if(isset($_POST['Delete'])){
    $id2 = $_POST['id'];

/**
**Updated your query to match the key rather than ID. `key`
**/
    $deleteQuery= "DELETE FROM tblfriends WHERE `key` = '$id2' ";
    if( mysqli_query($conn,$deleteQuery)){
        echo "Deleted";
    }else{
        echo "Not Deleted";
    }
};

为什么mysqli查询返回false？

2 个答案: