Question

我试图允许用户通过向上或向下移动来重新排序mySQL表中的行。我有2页允许这个工作。它不起作用，老实说我不知道从哪里开始。我收到此错误：Fatal error: Call to undefined method Database::GetAll() in /home/www/thetotempole.ca/phptester/moveupdown.php on line 35这里是我的php：

moveupdown.php

<?php 

class Database extends mysqli {
    function __construct() {
        parent::__construct("","","","");
        if (mysqli_connect_errno()) {
            throw new Exception(mysqli_connect_error(),   
            mysqli_connect_errno());
        } 
    } 
}

include 'connect.php'; 
$db = new Database(); 

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

    extract($_POST); 
    //determine what direction in relation to $_POST['position'] 
    $otherpos = $do=='⇑'? $position-1:$position+1; 
    //get the two ID that should change order place 
    $sql = "SELECT id FROM employees WHERE emp_id=$position"; 
    $posid = $db->GetRow($sql); 
    $sql = "SELECT id FROM employees WHERE emp_id=$otherpos"; 
    $other = $db->GetRow($sql); 
    //change place for those two 
    $sql = "UPDATE employees SET position=$otherpos WHERE id=$posid->id"; 
    $db->Query($sql); 
    $sql = "UPDATE forums SET position=$position WHERE id=$other->id"; 
    $db->Query($sql); 

}else{ 
    // make sure all forums positions are numbered 1,2,3,4,5 etc. 
    $sql = "SELECT id FROM employee ORDER BY position"; 
    $forums = $db->GetAll($sql); 
    foreach($forums AS $f){ 
        $items[] = $f->id;     
    } 
    foreach($items AS $k=>$id){ 
        $k++; 
        $sql = "UPDATE employee SET position=$k WHERE id=$id"; 
        $db->Query($sql); 
    } 
} 

//get the number of forums to display 
$sql = "SELECT COUNT(*) AS max FROM employee"; 
$pos = $db->GetRow($sql); 
//get them by order of position 
$sql = "SELECT * FROM employee ORDER BY position"; 
$forums = $db->GetAll($sql); 

//display with up/down arrows for each 
//except first forum has only down-arrow 
//and last forum has only up-arrow 
$out ='<table>'."\n"; 
foreach($forums as $f){ 
    $out .= '<form method="post">'; 
    $out .= '<input type="hidden" name="position" value="'.$f->position.'">'."\n".'<tr>'; 
    if($f->position != 1) 
        $out .= '<td><input type="submit" name="do" value="⇑"></td>';  //up 
    else $out .= '<td></td>'; 
    if($f->position != $pos->max) 
        $out .= '<td><input type="submit" name="do" value="⇓"></td>';  //down 
    else $out .= '<td></td>'; 
    $out .= '<td>'.$f->name.'</td></tr>'."\n"; 
    $out .= '</form>'."\n"; 
} 
$out .= '</table>'; 
echo $out; 

?>

connect.php

<?
//the example of MySQL database connection
//connect.php
$continued = mysql_connect("","","","");
if ($continued) {
    echo ("Connection is succeed");
} else {
    echo ("Connection is fail");
}
?>

Answer 1

这意味着Database类中没有GetAll方法，或者方法签名不同。在这种情况下，我没有在mysqli类中看到任何这些方法。检查docs of the mysqli class是否有效方法。

由于你扩展了你的Database类（不像其他评论者提到的那样，你可以只使用new mysqli()类），你继承了列出的所有mysqli方法在上面的链接。其中有一个mysqli方法，用于将SQL查询发送到服务器。所以你的代码部分看起来应该像（我不是在这里检查你的逻辑，只是替换与数据库相关的东西）：

query

}

PHP数据库类不起作用

1 个答案: