Question

情况如下：

我在PHP中有这个数组，用于保存查询结果中的数据：

$returnData = array(
    'ID'                 => '',
    'NAME'               => '', 
    'DESCRIPTION'        => '',
    'STATUS'             => '',
    'STATUS_DESCRIPTION' => '',
    'LOCATION'           => '',
    '_ERROR'             => ''
);

我稍后执行一个查询，我想填充这个数组：

$fetchResourceQuery = sprintf('SELECT  RESOURCEID, RESOURCENAME, RESOURCEDESCRIPTION,
                                           T3.RESOURCELOCATIONNAME, T2.RESOURCESTATUSNAME,
                                           T2.RESOURCESTATUSDESCRIPTION
                                   FROM resource          T1
                                   JOIN resource_status   T2
                                   ON T1.RESOURCESTATUSID = T2.RESOURCESTATUSID
                                   JOIN resource_location T3
                                   ON T1.RESOURCELOCATIONID = T3.RESOURCELOCATIONID');

$resultSet = $DB->query($fetchResourceQuery);
        if($resultSet){
            while($row = $resultSet->fetch_assoc()){
                $returnData['ID']                   = $row['RESOURCEID'];
                $returnData['NAME']                 = $row['RESOURCENAME'];
                $returnData['DESCRIPTION']          = $row['RESOURCEDESCRIPTION'];
                $returnData['STATUS']               = $row['RESOURCESTATUSNAME'];
                $returnData['STATUS_DESCRIPTION']   = $row['RESOURCESTATUSDESCRIPTION'];
                $returnData['LOCATION']             = $row['RESOURCELOCATIONNAME'];
            }
            $json_data = json_encode($returnData, JSON_UNESCAPED_SLASHES);
            echo $json_data;
        }

到目前为止这很好。问题是，我认为在保存数据的同时只有一个循环，然后由于某种原因覆盖在数组内部，输出是这样的：

{"ID":"456","NAME":"Rack con Televisor #1","DESCRIPTION":"Televisor Sharp Aquos con Laptop, armado en 2011.","STATUS":"Active","STATUS_DESCRIPTION":"Can be reserved.","LOCATION":"First Floor, High School","_ERROR":""}

这很糟糕，因为数据库中有5个资源，只存储其中一个资源。

如何使数组存储多个值，如josn对象？显然，我现在正在努力的灵魂不起作用。

Answer 1

$save = array();
while($row = $resultSet->fetch_assoc()){
    $returnData['ID']                   = $row['RESOURCEID'];
    $returnData['NAME']                 = $row['RESOURCENAME'];
    $returnData['DESCRIPTION']          = $row['RESOURCEDESCRIPTION'];
    $returnData['STATUS']               = $row['RESOURCESTATUSNAME'];
    $returnData['STATUS_DESCRIPTION']   = $row['RESOURCESTATUSDESCRIPTION'];
    $returnData['LOCATION']             = $row['RESOURCELOCATIONNAME'];
    array_push($save, $returnData);
}
$json_data = json_encode($save, JSON_UNESCAPED_SLASHES);
echo $json_data;

Answer 2

您确实在每次迭代时覆盖相同的对象。你想要的是这样的：

$returnData = array(); // Make it an array

// ...
// ...

while($row = $resultSet->fetch_assoc()){
    $item = array();
    $item['ID']                   = $row['RESOURCEID'];
    $item['NAME']                 = $row['RESOURCENAME'];
    $item['DESCRIPTION']          = $row['RESOURCEDESCRIPTION'];
    $item['STATUS']               = $row['RESOURCESTATUSNAME'];
    $item['STATUS_DESCRIPTION']   = $row['RESOURCESTATUSDESCRIPTION'];
    $item['LOCATION']             = $row['RESOURCELOCATIONNAME'];

    $returnData[] = $item; // Add new item to array
}

Answer 3

使用此：

$returnData = array(); // Simply do this. Don't need all index fields
$index = 0;
while($row = $resultSet->fetch_assoc()){
    $returnData[$index] = array("ID"=>$row['RESOURCEID'],
    "NAME"=>$row['RESOURCENAME'],"DESCRIPTION"=>$row['RESOURCEDESCRIPTION'],
    "STATUS"=>$row['RESOURCESTATUSNAME'],"STATUS_DESCRIPTION"=>$row['RESOURCESTATUSDESCRIPTION'],
    "LOCATION"=>$row['RESOURCELOCATIONNAME']);
    $index++;
}
echo json_encode($returnData, JSON_UNESCAPED_SLASHES);

Answer 4

$newArray = array();
while($row = $resultSet->fetch_assoc()){
            $returnData['ID']                   = $row['RESOURCEID'];
            $returnData['NAME']                 = $row['RESOURCENAME'];
            $returnData['DESCRIPTION']          = $row['RESOURCEDESCRIPTION'];
            $returnData['STATUS']               = $row['RESOURCESTATUSNAME'];
            $returnData['STATUS_DESCRIPTION']   = $row['RESOURCESTATUSDESCRIPTION'];
            $returnData['LOCATION']             = $row['RESOURCELOCATIONNAME'];

          array_push($newArray, $returnData);
        }

将mysql查询结果保存到PHP数组中

4 个答案: