SQL问题-php教程

回复讨论(解决方案)

... where instr('31971157310000016346802385782010',字段)>0

$val='634680238578';
$sql=“select * from tablename where fieldname like '%$val%'”;

不好意思，看错了.....

用 like 也是一样
.... ‘31971157310000016346802385782010’ like concat('%', 字段, '%')