我很好奇是否有可能使用PDO将值数组绑定到占位符。这里的用例试图传递一个值数组,以便与IN()条件一起使用。
我希望能够这样做:
<?php
$ids=array(1,2,3,7,8,9);
$db = new PDO(...);
$stmt = $db->prepare(
'SELECT *
FROM table
WHERE id IN(:an_array)'
);
$stmt->bindParam('an_array',$ids);
$stmt->execute();
?>
并让PDO绑定并引用数组中的所有值。
目前我正在做:
<?php
$ids = array(1,2,3,7,8,9);
$db = new PDO(...);
foreach($ids as &$val)
$val=$db->quote($val); //iterate through array and quote
$in = implode(',',$ids); //create comma separated list
$stmt = $db->prepare(
'SELECT *
FROM table
WHERE id IN('.$in.')'
);
$stmt->execute();
?>
这当然是工作,但只是想知道是否有一个内置的解决方案,我错过了?
当你有其他参数时,你可以这样做:
$ids = array(1,2,3,7,8,9);
$db = new PDO(...);
$query = 'SELECT *
FROM table
WHERE X = :x
AND id IN(';
$comma = '';
for($i=0; $i<count($ids); $i++){
$query .= $comma.':p'.$i; // :p0, :p1, ...
$comma = ',';
}
$query .= ')';
$stmt = $db->prepare($query);
$stmt->bindValue(':x', 123); // some value
for($i=0; $i<count($ids); $i++){
$stmt->bindValue(':p'.$i, $ids[$i]);
}
$stmt->execute();
稍微编辑一下施纳勒的代码
<?php
$ids = array(1, 2, 3, 7, 8, 9);
$inQuery = implode(',', array_fill(0, count($ids)-1, '?'));
$db = new PDO(...);
$stmt = $db->prepare(
'SELECT *
FROM table
WHERE id IN(' . $inQuery . ')'
);
foreach ($ids as $k => $id)
$stmt->bindValue(($k+1), $id);
$stmt->execute();
?>
//implode(',', array_fill(0, count($ids)-1), '?'));
//'?' this should be inside the array_fill
//$stmt->bindValue(($k+1), $in);
// instead of $in, it should be $id
你可以这样转换:
$stmt = $db->prepare('SELECT * FROM table WHERE id IN('.$in.')');
在此:
$stmt = $db->prepare('SELECT * FROM table WHERE id IN(:id1, :id2, :id3, :id7, :id8, :id9)');
然后用这个数组执行它:
$stmt->execute(array(
:id1 =>1, :id2 =>2, :id3 =>3, :id7 =>7, :id8 =>8, :id9 => 9
)
);
因此:
$in = array();
$consultaParam = array();
foreach($ids as $k => $v){
$in[] = ':id'.$v;
$consultaParam[':id'.$v] = $v;
}
最后的代码:
$ids = array(1,2,3,7,8,9);
$db = new PDO(...);
$in = array();
$consultaParam = array();
foreach($ids as $k => $v){
$in[] = ':id'.$v;
$consultaParam[':id'.$v] = $v;
}
$stmt = $db->prepare(
'SELECT *
FROM table
WHERE id IN('.$in.')'
);
$stmt->execute($consultaParam);
您首先在查询中设置“?”的个数,然后由一个“for”发送参数
像这样:
require 'dbConnect.php';
$db=new dbConnect();
$array=[];
array_push($array,'value1');
array_push($array,'value2');
$query="SELECT * FROM sites WHERE kind IN (";
foreach ($array as $field){
$query.="?,";
}
$query=substr($query,0,strlen($query)-1);
$query.=")";
$tbl=$db->connection->prepare($query);
for($i=1;$i<=count($array);$i++)
$tbl->bindParam($i,$array[$i-1],PDO::PARAM_STR);
$tbl->execute();
$row=$tbl->fetchAll(PDO::FETCH_OBJ);
var_dump($row);