我尝试使用下面的PHP从基于通过URL传递给它一个参数一个MySQL表中提取信息。页面的URL是job.php?id=2
,其中最后的数字2是要读取其数据的行的主键值。我能够echo
的$job_id
价值,但是当我试图在mysql_query()
功能,使用它,根本就没有数据返回。任何人都可以看到我要去哪里吗?问题与MySQL查询
<html>
<head>
<title></title>
</head>
<body>
<?
$job_id = $_GET["id"];
echo $job_id; // This successfully prints the Job_ID
$query = "SELECT *
FROM job JOIN (employer, address) ON
(
job.Employer_id_job = employer.Employer_ID
AND job.Address_id_job = address.Address_ID
)
WHERE Job_ID = $job_id";
if (!$query)
{
die ('Invalid query: ' .mysql_error());
}
$result = mysql_query($query);
$row = $job_id - 1;
$job_title = mysql_result ($result, $row, "Title");
$date_posted = mysql_result ($result, $row, "Date_posted");
$application_deadline = mysql_result ($result, $row, "Application_deadline");
$description = mysql_result ($result, $row, "Description");
$company = mysql_result ($result, $row, "Company_name");
$city = mysql_result ($result, $row, "City");
?>
<? echo $job_title; ?>
</body>
</html>