dtbrd80422 2016-04-10 07:28
浏览 27

PHP MySQL请求在使用变量时不起作用

the MySQL request works fine when I do it like this:

$sql1 = 'SELECT * FROM user WHERE username LIKE "test"';

but when I use a variable like this, it gives an error:

$username = $_POST["username"];
$sql1 = 'SELECT * FROM user WHERE username LIKE'.$username;

Error:

Fatal error: Call to a member function fetch_assoc() on boolean...

var_dump of $username gives:

 string(4) "test"

Thanks for your help!

Full code so far:

$username = $_POST["username"];
var_dump($username);

include 'data.php';
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

$sql1 = 'SELECT * FROM user WHERE username = "'.$username.'"';
$result1 = $conn->query($sql1);
$row1 = $result1->fetch_assoc();

echo $row1["ID"];
  • 写回答

2条回答 默认 最新

  • douxue7196 2016-04-10 07:30
    关注

    Because your code is incorrect:-> use this.... $sql1 = 'SELECT * FROM user WHERE username LIKE"'.$username.'"';

    评论

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