Php Mysql列未返回正确的值
我试图从一个特定的专栏中得到一门课程的剩余学额。但是,结果返回4,这是不正确的。它应该是15 代码如下:Php Mysql列未返回正确的值,php,mysql,loops,foreach,Php,Mysql,Loops,Foreach,我试图从一个特定的专栏中得到一门课程的剩余学额。但是,结果返回4,这是不正确的。它应该是15 代码如下: $courseID = $_GET['id']; $result = mysql_query("SELECT * FROM course_dates WHERE id = '$courseID'"); $row = mysql_fetch_row($result); foreach ($row as $record) { $placesLeft = $record['pla
$courseID = $_GET['id'];
$result = mysql_query("SELECT * FROM course_dates WHERE id = '$courseID'");
$row = mysql_fetch_row($result);
foreach ($row as $record) {
$placesLeft = $record['places'];
}
echo $placesLeft;
很明显,我是个彻头彻尾的傻瓜吗
编辑:$row的var_转储:
array(7) {
[0]=> string(1) "2"
[1]=> string(1) "2"
[2]=> string(56) "Essential leadership and mangement skills in the new nhs"
[3]=> string(6) "London"
[4]=> string(10) "2012-12-15"
[5]=> string(2) "15"
[6]=> string(3) "450"
}
我相信您的id是主键/唯一键 因此,您将只得到一行作为结果集 因此,使用:
$row['places'];
足够了,不要再重复它尝试以下方法: 使用
$courseID=“1”代码>和1
和将一个和西班牙
放在各自生成的列中:
1
放置一个
西班牙
您正在循环以获取val,并在循环外进行回声!!移动回声$placesLeft;例如:你什么都没做是什么意思?没有输出?是的,没有输出ID uprint\u r()
it??print\u r什么都不做,var\u dump=NULL
<?php
$DB_HOST = "xxx";
$DB_NAME = "xxx";
$DB_PASS = "xxx";
$DB_USER = "xxx";
$db = new mysqli($DB_HOST, $DB_USER, $DB_PASS, $DB_NAME);
if($db->connect_errno > 0) {
die('Connection failed [' . $db->connect_error . ']');
}
// $courseID = $_GET['id'];
$courseID = "1"; // my own test. Column set to (INT)
$result = mysqli_query($db,"SELECT * FROM course_dates WHERE id = '$courseID'");
while($row = $result->fetch_assoc()) {
// my own example columns. Change to suit
echo $row['id'];
echo "<br>";
echo $row['places'];
echo "<br>";
echo $row['city'];
}
mysqli_close($db);