我正在尝试从特定列中获取课程中剩余的位置数。但是,结果是返回 4 ,这是不正确的。它应该 15 。
以下是代码:
$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 $placesLeft;
很明显,我是一个完整的菜鸟吗?
编辑:$ row的var_dump:
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"
}
答案 0 :(得分:1)
我相信您的ID是主要/唯一密钥
所以你只能得到一行作为结果集
所以使用:
$row['places'];
足够不再重复它
答案 1 :(得分:1)
请尝试使用此方法:
使用$courseID = "1";
并在其各自的列中生成1
和Place ONE
以及Spain
:
1
放置一个
西班牙
<?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);
使用$courseID = "2";
并在其各自的列中生成2
和Place TWO
以及LONDON
:
2
放置两个
伦敦