4

我有一些这样的 php 代码:

$row = mysql_fetch_array ( mysql_query("SELECT * FROM `tblFacilityHrs` WHERE `uid` = '$uid'")); 

我现在正在尝试将其转换mysqli_fetch_array为此处所示的http://php.net/manual/en/mysqli-result.fetch-array.php(示例#1 面向对象的样式)

我不确定示例中的“$result”是什么意思。

到目前为止,这是我将代码转换为的内容:

<?php 
include('../config.php'); 
if (isset($_GET['uid']) ) { 
$uid = $_GET['uid'];
$id = $_GET['id'];  
if (isset($_POST['submitted'])) { 
foreach($_POST AS $key => $value) { $_POST[$key] = mysqli_real_escape_string($value); } 

//Query for tblFacilityHrs
$sql = " UPDATE tblFacilityHrs SET `title`='{$_POST['title']}',`description`='{$_POST['description']}' WHERE `uid` = '$uid' "; 
$result = $mysqli->query($sql) or die($mysqli->error);

//Query for tblFacilityHrsDateTimes
$sql2 = "UPDATE tblFacilityHrsDateTimes SET `startEventDate`='{$_POST['startEventDate']}',`endEventDate`='{$_POST['endEventDate']}', `startTime`='{$_POST['startTime']}',`endTime`='{$_POST['endTime']}',`days`='{$_POST['days']}',`recurrence`='{$_POST['recurrence']},`finalDate`='{$_POST['finalDate']}' WHERE `id` = '$id' "; print $sql2;
$result2 = $mysqli->query($sql2) or die($mysqli->error);

echo ($mysqli->affected_rows) ? "Edited row.<br />" : "Nothing changed. <br />"; 
echo "<a href='list.php'>Back</a>";
} 
$row = $result->fetch_array($mysqli->query("SELECT * FROM `tblFacilityHrs` WHERE `uid` = '$uid'"));
$row2 = $result2->fetch_array($mysqli->query("SELECT * FROM `tblFacilityHrsDateTimes` WHERE `id` = '$id'"));
?>

我在学习mysqli时可能有很多问题,但现在它会出错

致命错误:调用成员函数 fetch_array()

4

2 回答 2

5

UPDATE查询不返回结果对象,它返回TRUE(假设成功)。然后你试图打电话TRUE->fetch_array(),这显然是行不通的。

现在,为了做你真正想做的事,试试这个:

$row = $mysqli->query("SELECT.....")->fetch_array();
于 2012-05-09T14:28:27.060 回答
1

看起来您正在尝试使用旧的结果对象(更新的结果)来获取新查询的结果。您需要先运行新查询,将其结果分配给对象,然后从对象中获取结果。查看重新编写的脚本的最后三行:

$facilityHoursQueryResult = $mysqli->query("SELECT * FROM `tblFacilityHrs` WHERE `uid` = '$uid'");
$facilityHoursDateTimesQueryResult = $mysqli->query("SELECT * FROM `tblFacilityHrsDateTimes` WHERE `id` = '$id'");

$facilityHoursRow = $facilityHoursQueryResult == FALSE ? NULL : $facilityHoursQueryResult->fetch_array();
$facilityDateTimesRow = $facilityHoursDateTimesQueryResult == FALSE ? NULL : $facilityHoursDateTimesQueryResult->fetch_array();
?>

这是一个包含其他结果对象方法的有用页面: http ://www.php.net/manual/en/class.mysqli-result.php

于 2012-05-09T14:35:43.717 回答