我有这个查询在phpmyadmin中工作,但在php中不起作用。除此之外还有其他选择吗?
$servername = "localhost";
$username = "user";
$password = "password";
$dbname = "dbname";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {die("Connection failed: " . $conn->connect_error);}
$champname = 'name';
$sql = "SELECT column1, column2, column3, column4, column5
FROM champions
WHERE champion = '$champname' ORDER BY ID DESC LIMIT 1";
$result = $conn->query($sql);
$row = $result->fetch_assoc();
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我想'$champname'
是问题所在。
答案 0 :(得分:1)
试试这个
$sql = "SELECT column1, column2, column3, column4, column5 FROM champions WHERE champion = $champname ORDER BY ID DESC LIMIT 1";
或
$sql = "SELECT column1, column2, column3, column4, column5
FROM champions
WHERE champion = `$champname` ORDER BY ID DESC LIMIT 1";
答案 1 :(得分:0)
$ sql =“SELECT column1,column2,column3,column4,column5 来自冠军 WHERE冠军='“。$ champname。”'ORDER BY ID DESC LIMIT 1“;