根据另一个表的结果导入到mysql

时间:2018-07-16 19:57:09

标签: php mysql json

我希望将JSON数据导入到表中,但我只想根据另一个表中的内容来获取字段。

这是我的示例数据库:

Array: [a, b, c]

因此,然后基于该映射(即街道地址为字段11),我希望能够解析POSTed JSON并获取这些字段并将其插入到另一个表中。

CREATE TABLE `fields` (
`id` int(11) unsigned NOT NULL AUTO_INCREMENT,
`fieldname` varchar(50) DEFAULT NULL,
`encompassid` varchar(50) DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4;

INSERT INTO `fields` (`id`, `fieldname`, `fileid`)
VALUES
(1,'streetaddress','11'),
(2,'city','12');

当前我的代码返回正确的查询,除了变量没有来自JSON的数据,它们只是一个字符串。

CREATE TABLE `testfile` (
`id` int(11) unsigned NOT NULL AUTO_INCREMENT,
`streetname` varchar(100) DEFAULT NULL,
`city` varchar(50) DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=29 DEFAULT CHARSET=utf8mb4;

这是示例JSON:

$con=mysqli_connect($dbhost,$dbuser,$dbpassword,$db);
$result = mysqli_query($con,"SELECT * FROM fields");
while($row = mysqli_fetch_assoc($result)) {
    $data[] = $row;
    $items = array();
    $itemsV = array();
        $sfv = '';
    foreach ($data as $obj1) {
        if($sfv != "\$".$obj1['fieldname']." = \$obj2->{'".$obj1['fileid']."'};"){
            $sfv .= "\$".$obj1['fieldname']." = \$obj2->{'".$obj1['fileid']."'};";
        }

        $items[] .= $obj1['fieldname'];
        $itemsV[] .= "\$".$obj1['fieldname'];
    }

}


$string = file_get_contents('php://input');
$data2 = json_decode($string);
foreach ($data2 as $obj2) {

        $values  = implode(',', $itemsV);

        eval("\$sfv;");

        $columns = implode(", ",$items);

        $insert = "REPLACE INTO testfile ($columns) VALUES ($values);";

        echo $insert;

}

当前输出看起来像这样:{ "serializedExport": null, "format": 1, "fields": { "2": "100000", "3": "4.500", "4": "360", "5": "760.03", "6": "", "8": "", "9": "", "10": "", "11": "123 HOPE STREET", "12": "Philadelphia", "13": "", "14": "PA", "15": "19119", "16": "1", "17": "" } } 我在哪里REPLACE INTO fields_testloan (streetaddress, city) VALUES ($streetaddress,$city);

这可能吗?

1 个答案:

答案 0 :(得分:2)

在读取fields表时,创建一个从字段ID映射到变量名的关联数组。

$fields = array();
while ($row = mysqli_fetch_assoc($result)) {
    $fields[$row['fileid']] = $row['fieldname'];
}

然后在处理JSON时,创建字段名称和值的数组。使用json_decode()的第二个参数,可以得到一个关联数组而不是对象,因此可以使用foreach对其进行循环。

$string = file_get_contenst("php://input");
$data2 = json_decode($string, true);

$colnames = array();
$colvalues = array();
foreach ($data["fields"] as $fieldid => $fieldvalue) {
    if (isset($fields[$fieldid])) {
        $colnames[] = $fields[$fieldid];
        $colvalues[] = "'" . mysqli_real_escape_string($con, $fieldvalue) . "'";
    }
}

$cols_string = implode(",", $colnames);
$vals_string = implode(",", $colvalues);
$sql = "REPLACE INTO testfile ($cols_string) VALUES ($vals_string)";