Question

已经忙碌了一段时间但是在删除一行后仍然无法重新加载页面。

行删除工作正常，但只有在我重新加载页面后才会显示。一直在移动一些代码，以便在按下提交时读取但没有运气。任何想法？

 <?php
mysql_connect("localhost", "root", "") or die(mysql_error()) ; 
mysql_select_db("booking") or die(mysql_error()) ;

$resultselect = mysql_query("SELECT * FROM test_mysql ") or die(mysql_error()); 
?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
    "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> 

<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en"> 

    <head> 
        <meta http-equiv="Content-Type" content="text/html; charset=UTF-8" /> 
        <title>row delete</title> 
        <link rel="stylesheet" type="text/css" href="" /> 
    </head> 

    <body>
<?php 
    if(isset($_POST['done'])){
    $done = implode(',', $_POST['done']);
    $deletequery = "DELETE FROM test_mysql WHERE id IN ($done)";
    $resultdelete = mysql_query($deletequery) or die(mysql_error());
    }

    ?>

    <form method="post" action="">
    <?php while($row = mysql_fetch_array($resultselect)){?>
        <input type="checkbox" name="done[]" 
    id="<?php echo $row['id'] ?>"
    value="<?php echo $row['id'] ?>">
    <label for="<?php echo $row['id'] ?>">
    <?php echo $row['name'] . ' ' . $row['lastname']. ' '.$row['email']; ?>
    </label>
    <br />
    <?php } ?>
    <input type ="submit" value ="submit">
    </form>
    <?php 

    if($resultdelete) {
        echo 'Item(s) deleted from list.';
    }
    ?>

    <pre>
    <?php print_r($_POST) ?>
     </pre>
    </body>
</html>

Answer 1

我假设，因为您提供的代码是您的应用程序代码的片段，$resultselect在此部分执行之前就已设置好。

这里重要的部分是MySQL查询的顺序。如果您在SELECT行之前DELETE行，那么当您运行mysql_fetch_array()时，您仍会看到已删除的行

在SELECT查询后移动DELETE查询，一切正常。

Answer 2

看起来你在delete语句之前的代码段中得到$ row的结果集。如果将$ deletequery代码放在页面顶部，则可以删除相应的数据记录，然后获取新的结果集，然后将其显示给用户。

无论您的循环是什么查询

while($row = mysql_fetch_array($resultselect))
{
....
}

需要在

之后运行

$resultdelete = mysql_query($deletequery) or die(mysql_error());

语句。

编辑要运行的完整代码：

<?php
    mysql_connect("localhost", "root", "") or die(mysql_error()) ; 
    mysql_select_db("booking") or die(mysql_error()) ;

    if(isset($_POST['done']))
    {
        $done = implode(',', $_POST['done']);
        $deletequery = "DELETE FROM test_mysql WHERE id IN ($done)";
        $resultdelete = mysql_query($deletequery) or die(mysql_error());
    }
    $resultselect = mysql_query("SELECT * FROM test_mysql ") or die(mysql_error()); 
?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
    "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> 

<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en"> 

    <head> 
    <meta http-equiv="Content-Type" content="text/html; charset=UTF-8" /> 
    <title>row delete</title> 
    <link rel="stylesheet" type="text/css" href="" /> 
    </head> 

    <body>
    <form method="post" action="">
    <?php while($row = mysql_fetch_array($resultselect)){?>
    <input type="checkbox" name="done[]" 
    id="<?php echo $row['id'] ?>"
    value="<?php echo $row['id'] ?>">
    <label for="<?php echo $row['id'] ?>">
    <?php echo $row['name'] . ' ' . $row['lastname']. ' '.$row['email']; ?>
    </label>
    <br />
    <?php } ?>
    <input type ="submit" value ="submit">
    </form>
    <?php 

    if($resultdelete) {
    echo 'Item(s) deleted from list.';
    }
    ?>

    <pre>
    <?php print_r($_POST) ?>
     </pre>
    </body>
</html>

从sql中删除行后重新加载页面

2 个答案: