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php like 查询

WBOY
Release： 2016-06-23 13:32:35
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$SA_IP=北京市;

$citya =北京市,天津市;
$cityb =上海市,深圳市;
$citys =$citya+$cityb;

if($SA_IP==$citys){// 表示找到匹 ?>
       
    
  else{
  ?>
  
  }?>

就是当$SA_IP=北京市;，条件真。当$SA_IP=天津市;，条件真。 当$SA_IP=深圳市; ，条件真。

SQL有个like %% 可以模糊查询，就是我现在用PHP要怎么样去实现。


回复讨论(解决方案)

strpos 

header ( 'Content-Type:text/html;charset=utf-8' );$SA_IP = '北京市';$citys = '北京市,天津市,上海市,深圳市';if (strpos ( $citys, $SA_IP ) !== false) {	echo '存在';} else {	echo '不存在';}
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strpos楼上正解 

header ( 'Content-Type:text/html;charset=utf-8' );$SA_IP = '北京市';$citys = '北京市,天津市,上海市,深圳市';if (strpos ( $citys, $SA_IP ) !== false) {	echo '存在';} else {	echo '不存在';}
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$citya =北京市,天津市;
$cityb =上海市,深圳市;
$citys =$citya+$cityb;
$citys的数据是这$citya $cityb 相加的数据，能相加吗 

header ( 'Content-Type:text/html;charset=utf-8' );$SA_IP = '北京市';$citys = '北京市,天津市,上海市,深圳市';if (strpos ( $citys, $SA_IP ) !== false) {	echo '存在';} else {	echo '不存在';}
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谢谢 使用 $citys =$citya.$cityb; 就相加了
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php like 查询
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