select * like with masked compare


select * like with masked compare

我必须在数据库中搜索类似的$string s。A'?'在引用中,String表示任何字符,因此必须忽略此字符,但所有其他字符必须匹配。

参考CCD_ 2将与CCD_ 3或CCD_。mySQL"Select…"看起来怎么样?

$string = 'abcd?efg';    
$row = $mysqli->query("SELECT * FROM $table WHERE String LIKE '$string'");
$result = $row->fetch_assoc();
print_r( json_encode($result) );

LIKE中的1个字符通配符是_,因此使用str_replace?替换为_

$like_string = str_replace('?', '_', $string);
$stmt = $mysqli->prepare("SELECT * FROM $table WHERE String LIKE ?");
$stmt->bind_param("s", $like_string);
$stmt->execute();

您可以使用这样的正则表达式:

$string = 'abcde?efg';
$searchGroup = explode('?', $string);    
$row = $mysqli->query("SELECT * FROM $table WHERE String REGEXP '^" . $searchGroup[0].".".$searchGroup[1]."$'"); // starts with abcde and ends with efg and allows one character between the groups.

这将在两个组之间获得一个字符,它可以是任何字符。来自文档:

"."匹配任何单个字符。