在 Jqgrid 中使用 SQL
Using SQL Like in Jqgrid
我正在使用 guriddo 中的 jqGrid,并希望动态创建一个包含 LIKE 的查询。我有以下代码:
$sql = "SELECT * FROM users";
if (isset($_SESSION['search']))
{
$temp = $_SESSION['search'];
$sql .= " WHERE users.name LIKE %".$temp."%";
}
$grid->SelectCommand = $sql;
但这给了我以下错误:
SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '%'Adam'% LIMIT 0, 0' at line 9
我认为您在 Like
周围缺少单引号 '
以下应该有效
$sql = "SELECT * FROM users";
if (isset($_SESSION['search']))
{
$temp = $_SESSION['search'];
$sql .= " WHERE users.name LIKE '%".$temp."%'";
}
$grid->SelectCommand = $sql;
你错过了一个报价。应该是
$sql .= " WHERE users.name LIKE '%".$temp."%'";
我正在使用 guriddo 中的 jqGrid,并希望动态创建一个包含 LIKE 的查询。我有以下代码:
$sql = "SELECT * FROM users";
if (isset($_SESSION['search']))
{
$temp = $_SESSION['search'];
$sql .= " WHERE users.name LIKE %".$temp."%";
}
$grid->SelectCommand = $sql;
但这给了我以下错误:
SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '%'Adam'% LIMIT 0, 0' at line 9
我认为您在 Like
'
以下应该有效
$sql = "SELECT * FROM users";
if (isset($_SESSION['search']))
{
$temp = $_SESSION['search'];
$sql .= " WHERE users.name LIKE '%".$temp."%'";
}
$grid->SelectCommand = $sql;
你错过了一个报价。应该是
$sql .= " WHERE users.name LIKE '%".$temp."%'";