php中select查询用like失败请教大神,看看我是哪里写错了吗?把like条件去除就可以运行

问题描述:

<?php header("content-Type: text/html; charset=utf-8");
include '../conn.php';
$page = $_GET['PAGE'];
$name = $_GET['NAME'];
$display = 10; //每页显示10个数 据
$start = ($page-1)*$display;
$queryResult = $connect->query("SELECT usedcar_autobrandid, usedcar_class, usedcar_name, usedcar_detail, usedcar_price, usedcar_yprice, usedcar_year, usedcar_lt,
usedcar_ltl, usedcar_amt, usedcar_register, usedcar_mileage, usedcar_colour, usedcar_add, usedcar_author, usedcar_tal, usedcar_sub, usedcar_display, usedcar_video,
usedcar_pic1, usedcar_pic2, usedcar_pic3, usedcar_pic4, usedcar_pic5, usedcar_pic6, usedcar_pic7, usedcar_pic8, usedcar_pic9, usedcar_pic10
FROM usedcar_detail WHERE usedcar_display = 1 AND usedcar_name = $name LIKE '%$name%' Order By usedcar_time Desc LIMIT $start,$display");
$result = array();
while ( $fetchData = $queryResult -> fetch_assoc()){
$result [] = $fetchData;
}
echo json_encode($result,JSON_UNESCAPED_UNICODE);
mysqli_close($connect);
?>

usedcar_name  LIKE '%$name%'

FROM usedcar_detail WHERE usedcar_display = 1 AND usedcar_name = $name LIKE '%" + $name + "%' Order By usedcar_time Desc LIMIT $start,$display");

注意是百分号,不是星号,而且字符串和变量的拼接你也没写对