Question

我有一个像'迪拜，沙迦'这样的字符串，我想把它变成'迪拜'，'沙迦'，

我使用ajax和my code

在url中传递值dubai，sharjah

$city=$_GET['city'];

$sql  = "SELECT * FROM filter where isdeleted = 0 ";

if ($city !="" && $city !="Empty" ){
     $sql.=" and twon in ('".$citydata."')";     

            }

当我打印查询时，就像这样

SELECT * FROM filter where isdeleted = 0 and twon in ('dubai,sharjah,')

但我想这样

SELECT * FROM filter where isdeleted = 0 and twon in ('dubai','sharja')

任何人都可以使用php指导我如何做到这一点。

Answer 1

以下是我接近它的方法。我将使用PDO而不是 mysqli ，因为尝试将数组放入mysqli_stmt::bind_param只是一种痛苦。

首先，创建一个城市数组，删除所有空值

$params = array_filter(explode(',', $city), function($c) {
    return !empty($c) && $c !== 'Empty';
});
$paramCount = count($params);

$query = 'SELECT * FROM filter where isdeleted = 0';

现在为准备好的语句生成一个占位符字符串。

if ($paramCount) {
    $placeholders = implode(',', array_fill(0, $paramCount, '?');
    // looks something like '?,?'

    $query .= " AND twon IN ($placeholders)";
}

现在，准备一份声明

// assuming you have a PDO instance in $pdo created with something like
// $pdo = new PDO('mysql:host=localhost;dbname=your_db;charset=utf8', 'username', 'password', [PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION]);

$stmt = $pdo->prepare($query);

执行并获取值：）

$stmt->execute($params);
$filters = $stmt->fetchAll(PDO::FETCH_ASSOC);

Answer 2

$cities = explode(",", $_GET['city']);

//escape!
foreach ($cities as $citykey => $city) {
    $cities[$citykey] = "'" . mysql_real_escape_string($city) . "'";
}

$sql = "SELECT * FROM `filter` where `isdeleted` = 0";
if (!empty($cities)) {
    $sql .= ' and `twon` in (' . implode(',', $cities) . ')';
}

Answer 3

另一种方法是使用FIND_IN_SET()。无需更改PHP代码。

$sql.=" and FIND_IN_SET(town, '".$citydata."')";

Answer 4

你必须在这里爆炸和崩溃。休息是最后字符串中额外,的问题。

$str = "dubai,sharjah,";
$citydata = implode("','",explode(',',rtrim($str,',')));
echo "'$citydata'";

test

Answer 5

在6个答案之后得到第7个

$sql.=" and twon in ('".str_replace(",","','",$citydata)."')";

Answer 6

你可以尝试爆炸字符串

$cityparts = explode(",",$city);

你可以使用

$cityparts[0] and $cityparts[1] in your query 

array explode ( string $delimiter , string $string [, int $limit ] )

您可以在[http://www.php.net/explode]

上找到更多信息

希望这会有所帮助!!

Answer 7

你可以这样做。

$string = 'dubai,sharjah';
$cities = explode(',', $string);

echo $cities[0]; //dubai
echo $cities[1]; //sharjah

Answer 8

试试这个

$str = "dubai,sharjah,";
$arr = explode(",", $str);
$arr_temp = array()
foreach($arr as $s)
{
   if($s!="")
   {
      $arr_temp[] = "'".$s."'";
   }
}

$new_str = implode(",", $arr_temp); 
echo $new_str;  // output 'dubai','sharjah'

现在你的Sql将是

$sql = "SELECT * FROM filter where isdeleted = 0 and twon in ($new_str) ";

Answer 9

你可以使用它在$city我修复了两个值，您可以在$_GET;

使用

$city = "dubai,sharjah,";
$query_string ="";
$words = explode(",",$city);
            for($i=0;$i<count($words);$i++){$query_string .= "'".$words[$i]."',";}
            $query_string = substr($query_string,0,strlen($query_string)-4);

然后使用您的查询

SELECT * FROM filter where isdeleted = 0 and twon in ($query_string)

Answer 10

if ($city !="" && $city !="Empty" )
{
     $city_exp = explode(',' $city);
     $sql .= " and twon in ('".implode("', '", $city_exp)."')"; 
}

我们在这里基本上做的是使用explode将两个值放在一个数组中，然后使用implode

分隔该数组中的每个项目

DEMO

如何使用php将字符串拆分为子字符串

10 个答案: