我有一个像这样的表[info]
+---+-------+---------------+
|id | cost | place |
+---+-------+---------------+
|1 | 2000 | Dhaka |
|2 | 1000 | Cox's Bazar |
+---+-------+---------------+
现在我使用此查询来显示这些数据
$a_place = $_POST['place'];
query = "SELECT * FROM info WHERE place = '$a_place'";
我在寻找达卡时工作正常,但它不适用于考克斯的巴扎尔。也许为此> '
现在我该怎么办?请帮忙!
答案 0 :(得分:1)
$a_place = str_replace($_POST['place'],"'","''");
query = "SELECT * from info WHERE place = '".$a_place."'";
创建表格和插入数据
正如您所看到的那样,数据就像您的一样。
如果我像我一样选择输出 true
答案 1 :(得分:0)
也许这有助于一个开始:
// prepare and bind
$a_place = $_POST['place'];
$stmt = $conn->prepare("SELECT country FROM info WHERE place = '?'");
$stmt->bind_param("s", $a_place);
$stmt->execute();
$stmt->store_result();
if($stmt->num_rows){
$stmt->bind_result($country);
$stmt->fetch();
$stmt->free_result();
echo $country;
};
?>
并看到这篇文章: Getting results of statement
(记得选择*是不好的做法)