我可以使用以下代码在PDO准备好的语句中成功实现IN子句.
in_array = array(1,2,3);
$in = str_repeat('?,', count($in_array) - 1) . '?';
$sql = "SELECT * FROM my_table WHERE my_value IN ($in)";
$stm = $db->prepare($sql);
$stm->execute($in_array);
$data = $stm->fetchAll();
如何为多个$in做同样的事情?例如,我尝试以下操作失败:
in_array1 = array(1,2,3);
$in1 = str_repeat('?,', count($in_array) - 1) . '?';
in_array2 = array(4,5,1);
$in2 = str_repeat('?,', count($in_array) - 1) . '?';
$sql = "SELECT * FROM my_table WHERE (my_value1 IN ($in1)) AND (my_value2 IN ($in2))";
$stm = $db->prepare($sql);
$stm->execute($in_array1,$in_array2);
$data = $stm->fetchAll();
我认为这与stm-> execute有关,但不确定,请帮助
解决方法:
SELECT * FROM my_table WHERE (my_value1 IN (?,?,?)) AND (my_value2 IN (?,?,?))
因此,您的执行用法不正确,为http://php.net/manual/en/pdostatement.execute.php.它只应传递其中包含值的一个数组.
An array of values with as many elements as there are bound parameters in the sql statement being executed. All values are treated as PDO::ParaM_STR.
我认为使用array_merge,http://php.net/manual/en/function.array-merge.php将使您完成尝试的工作
$stm->execute(array_merge($in_array1,$in_array2));
这样的执行相当于
$stm->execute(array(1,2,3,4,5,1));
这似乎是不正确的,因为数组配对现已消失,但占位符1(第一个问号)将映射为1,占位符4至4,依此类推.