Question

我尝试从合并表中的特定行获取存储在SQL中的数据

这是我到目前为止所尝试的

<?php
$id = $_GET['id'];

$getdetails = mysql_query("SELECT 
scholarship.scholarshipid,
scholarship.scholarshipname, 
scholarship.shortdescription, 
scholarship.scholarshippicture,

scholarshipdetail.fulldescription,
scholarshipdetail.degreetype,
scholarshipdetail.location,
scholarshipdetail.deadline
FROM scholarship, scholarshipdetail
WHERE scholarship.scholarshipid = scholarshipdetail.scholarshipid AND scholarship.scholarshipid = $id ");

$retval = mysql_query( $getdetails, $conn );
if(! $retval )
{
die('Could not get data: ' . mysql_error());
}

?>

id来自 theurl.php？id = IDNUMBER 但事实证明它无法获取数据。如何从PHP中的ID号指定的行中获取数据？

Answer 1

您已尝试对另一个mysql_query的结果执行mysql_query！

让我们假设您的SQL目前是正确的，并处理其余代码。首先，您需要使用MySQLi或PDO，因为mysql扩展名已弃用。所以在MySQLi中;

$mysqli = new mysqli('host', 'user', 'password', 'db'); // fill in your details

$id = $_GET['id'];
if($stmt = $mysqli->prepare("SELECT 
scholarship.scholarshipid,
scholarship.scholarshipname, 
scholarship.shortdescription, 
scholarship.scholarshippicture,
scholarshipdetail.fulldescription,
scholarshipdetail.degreetype,
scholarshipdetail.location,
scholarshipdetail.deadline
FROM scholarship, scholarshipdetail
WHERE scholarship.scholarshipid = scholarshipdetail.scholarshipid
AND scholarship.scholarshipid = ?")) {

    $stmt->bind_param('i',$id);
    $stmt->execute();
    $result = $stmt->get_result();
}
else {
    echo $mysqli->error;
}

while($row = $result->fetch_assoc()) {
    // do stuff
    // cols stored as $row['col_name'];
}

请注意?所准备的SQL语句中的$id。这是变量的占位符，然后与$stmt->bind_param('i',$id);绑定（i表示整数，s将用于字符串）。然后，您必须执行结果并获得结果集，然后才能对其执行任何操作。

如果您的SQL出错，则错误将输出到浏览器，查询将无法执行。

希望这有帮助。

从ID指定的合并表中获取特定的数据行

1 个答案: