Question

我想知道如何将变量合并到PHP语句中以检查表是否存在。由于某种原因，查询不接受变量。这就是我所拥有的：

<?php
$servername = "localhost";
$username = "***";
$password = "***";
$dbname = "stavacom_students";

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}

$id = "1";
echo $id;
$query101 = 'select 1 from "$id" LIMIT 1';
$val = mysqli_query($conn, $query101);
if($val !== FALSE){

echo "no";

} else { 
?>

what?

<?php
};
?>

Answer 1

这些陈述需要StoredProcedureQuery query = entityManager .createStoredProcedureQuery("post_comments") .registerStoredProcedureParameter(1, Long.class, ParameterMode.IN) .registerStoredProcedureParameter(2, Class.class, ParameterMode.REF_CURSOR) .setParameter(1, 1L); query.execute(); List<Object[]> postComments = query.getResultList(); $ id` LIMIT 1“;

Answer 2

您可以使用mysql以两种方式实现此目的，第一种是简单SELECT FROM类型的查询：

<?php
function mysql_table_exists(mysqli $conn, $table)
{
        $res = @$conn->query('SELECT 1 FROM `' . $conn->real_escape_string($table) . '`');

        if($res)
        {
                $res->free();

                return true;
        }

        return false;
}
?>

您也可以使用以下语句在INFORMATION_SCHEMA中执行检查来执行此操作：

SELECT 1 FROM information_schema.tables WHERE table_schema = '<database name>' AND table_name = '<table name>' LIMIT 1;

最后，您可以调用SHOW TABLES LIKE '<table name>';并解析匹配结果，类似于第一个示例中的概念。

Answer 3

使用双引号（“...查询...”）编辑查询

$query101 = 'select 1 from "$id" LIMIT 1';

进入

$query101 = "select 1 from '".$id."' LIMIT 1";

OR

 $query101 = "select 1 from $id LIMIT 1";

Answer 4

改变这一点，希望它有用

$query101 = 'select 1 from "$id" LIMIT 1';

到这个

$query101 = 'select 1 from "'.$id.'" LIMIT 1';

检查现有表使用变量

4 个答案: