【发布时间】：2018-09-11 19:40:34
【问题描述】：
我正在尝试使用 JSON 文件中的值更新数据库中的值：
代码：
$jsonData = file_get_contents($jsonFile);
$data = json_decode($jsonData, true);
//check if hospital exist
$name = explode(' ',trim($data['organisationUnits']['organisationUnit']['name']));
// echo $name[0];
$query = Hospital::where('h_name', 'LIKE' , '%' . $data['organisationUnits']['organisationUnit']['name'] . '%')->first();
if($query){
// echo "\n yupo";
$h_id = $query->id;
$h_slug = $query->h_slug;
$nr_orgUnit = $query->nr_orgUnit;
// echo $nr_orgUnit;
$updateHospital = Hospital::find($h_id);
$updateHospital->h_name = $data["organisationUnits"]["organisationUnit"]["name"];
$updateHospital->h_short_name = $data["organisationUnits"]["organisationUnit"]["shortName"];
$updateHospital->h_code = $data["organisationUnits"]["organisationUnit"]["code"];
$updateHospital->h_opening_date = $data["organisationUnits"]["organisationUnit"]["openingDate"];
$updateHospital->h_closed_date = $data["organisationUnits"]["organisationUnit"]["closedDate"];
$updateHospital->h_active = $data["organisationUnits"]["organisationUnit"]["active"];
$updateHospital->h_comment = $data["organisationUnits"]["organisationUnit"]["comment"];
$updateHospital->h_geo_code = $data["organisationUnits"]["organisationUnit"]["geoCode"];
$updateHospital->h_last_updated = $data["organisationUnits"]["organisationUnit"]["lastUpdated"];
$updateHospital->save();
} else {
// echo 'error';
}
JSON 数据：
{"organisationUnits":{
"organisationUnit":{
"id":"01",
"uuid":{
},
"name":"Isagehe Dispensary",
"shortName":"Isagehe Dispensary ",
"code":"17-04-0118",
"openingDate":"1990-01-01",
"closedDate":{
},
"active":"true",
"comment":{
},
"geoCode":{
},
"lastUpdated":{
}
}
}
}
当我尝试运行代码时，我收到以下错误：
数组到字符串的转换（SQL：update `ag_hospitals` set `h_closed_date` = , `h_active` = true, `h_comment` = , `h_geo_code` = , `h_last_updated` = where `id` = 41）”
我哪里错了？
注意我也尝试过以下方式更新：
$updateHospital = Hospital::where('id', $h_id)->update([
'h_name' => $data['organisationUnits']['organisationUnit']['name'],
'h_short_name' => $data['organisationUnits']['organisationUnit']['shortName'],
'h_code' => $data['organisationUnits']['organisationUnit']['code'],
'h_opening_date' => $data['organisationUnits']['organisationUnit']['openingDate'],
'h_closed_date' => $data['organisationUnits']['organisationUnit']['closedDate'],
'h_active' => $data['organisationUnits']['organisationUnit']['active'],
'h_comment' => $data['organisationUnits']['organisationUnit']['comment'],
'h_geo_code' => $data['organisationUnits']['organisationUnit']['geoCode'],
'h_last_updated' => $data['organisationUnits']['organisationUnit']['lastUpdated']
]);
【问题讨论】：
-
$data["organisationUnits"]["organisationUnit"]["closedDate"]是字符串吗？试试看。 -
$data['organisationUnits']['organisationUnit']['comment']和$data['organisationUnits']['organisationUnit']['geoCode']是空对象，而不是字符串。您可能需要检查这些值是否为空，并写入一个空字符串或 null 而不是该值。 -
我已将它们设置为 NULL 并且有效，如何将对象转换为
$data['organisationUnits']['organisationUnit']['comment']和$data['organisationUnits']['organisationUnit']['geoCode']上的字符串？ -
对于空对象，只需使用
empty($data['organisationUnits']['organisationUnit']['comment']) ? '' : $data['organisationUnits']['organisationUnit']['comment']。当有数据时，您可能需要仔细检查结构以找出它是字符串、对象还是数组，然后从那里开始。
标签： php json laravel laravel-5