我想编辑一个JSON对象，因此使方法称为<div class="cards-container">
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</div>，该函数从数据库中找到特定的JSON alterJSON，然后根据参数更改array
所以我的问题是，正如您在此foreach循环中看到的那样，有一个key-values
(if statement)循环将打印foreach( $decoded as $index => $product ){
print_r('<pre>');
if( $product->$key == $item) {
echo ('found');
$decoded[ $index ]->$key = $value;
$json = json_encode( $decoded );
$stmt->execute([
':json' => $json,
':id' => $bid
]);
}
}
，并通过found查询在$JSON上进行一些更改，这是完美的工作，但是即使存在，JSON也只会找到JSON MySQL在同一个object or item中多于一个item，所以，我想在key-value中创建另一个条件，以检查唯一的statement
称为key-value，那么我该怎么做呢？我做了一些尝试却失败了。
我尝试过的
id <?php
public function AlterJSON( $bid=false, $item=false, $key=false, $Product, $id, $value=false ){
try{
if( $bid && $item && $key && $value && $Product && $this->connected === true ){
$stmt=$this->connection->prepare('select `items` from `bills` where `id`=:id');
$stmt->execute( [ ':id' => $bid ] );
$fetched = $stmt->fetchColumn();
$decoded = json_decode( $fetched );
$stmt=$this->connection->prepare('update `bills` set `items`=:json where id=:id');
foreach( $decoded as $index => $product ){
print_r('<pre>');
if( $product->$key == $item && $product->$id == $Product) {
echo ('found');
$decoded[ $index ]->$key = $value;
$json = json_encode( $decoded );
$stmt->execute([
':json' => $json,
':id' => $bid
]);
}
}
return true;
}
return false;
} catch( PDOException $e ){
return $this->errors === true ? $this->error( $e->getMessage() ) : false;
}
}
?>
function call:-答案 0 :(得分：0)
您是否尝试过将$stmt=$this->connection->prepare放入foreach循环中？
可以肯定的是，函数AlterJSON具有变量$Product，但是当您调用此AlterJSON时使用$ProductID时，是否可以同时使用$ProductID两个变量？ PHP对变量区分大小写。这样很好，但我认为可读性更好。
如果if( $product->$key中不存在$key，则$product中的其他技巧也会产生错误。是的，您可以尝试一下，但是如果替换项位于N循环中，则将失去进行此更新的可能性。我会在$product->$key之前安装保护，以确保密钥存在。