Question

这是我的json解码代码，解码后插入表中。我没有得到如何将implode值插入表和嵌套的json值。

include 'config.php'; /* datdabase connection*/
    class User {
public function update()
{
    $jsondata = file_get_contents('data.txt');
    $data = json_decode($jsondata, true);
$name=array();
    foreach($data as $item)
    {
$name[]=$item['v'];

        }
$val=implode(',',$name);
        echo"$val";
        $sql= mysql_query("INSERT INTO incident_main(inc_patientName,inc_patientAge,inc_patientGender,inc_patientMobile,inc_patientAddress,inc_patientWard,inc_type)  VALUES('..........') )");
        if(!$sql)
        {
            echo "Failed to insert the data into table";

    }
}}

$obj=new User(); 
$obj->update()

这是我的data.txt文件。这里“n”是表格列的名称，“v”是要插入的值。

{
    "user_id": "user123",
    "time_stamp": "epoch",
    "lat": 52.25,
    "lon": 85.45,
    "mainreport":    [{
         "n": "date_of_diagnosis",
         "v": "epoch"
        },
        {
         "n": "name",
         "v": "abc"
        },
        {
         "n": "age",
         "v": 20
        },
        {
         "n": "gender",
         "v": "male"
        },
        {
         "n": "address",
         "v": "some address"
        },
        {
         "n": "location_gis_Stamp",
         "v": "word-1"
        },
        {
         "n": "mobile_number",
         "v": "some number"
        },
        {
         "n": "type_of_application",
         "v": "other",
         "nested": [{
             "n": "other",
             "v": "reason"
         }]}]}

Answer 1

您的查询中有以下错误

表格名称和字段incident_main(inc_patientName,inc_patientAge,........
您在查询中使用了额外的)近值。 inc_treatment,inc_untraced) VALUES('..........') )"
用'value'括起数据库值，如果你只有整数值就没问题，所以你不需要添加''。

试试这样：

foreach($data as $item)
{
    $name[]=$item['v'];

    // Check and get Nested Values
    if( isset($item['nested']) && is_array($item['nested']) ) {
        foreach($item['nested'] as $val) {
            $name[] = $val['v'];
        }
    }

}


$val=implode("','",$name);
// epoch', 'abc', '20', 'male', 'some address', 'word-1', 'some number', 'other', 'reason
$sql= mysql_query("INSERT INTO incident_main (inc_patientName,inc_patientAge,inc_patientGender,inc_patientMobile,inc_patientAddress,inc_patientWard,inc_type,inc_diagnosis,inc_recurrence,inc_hospitalized,inc_treatment,inc_untraced) 
        VALUES('".$val."') ");

请使用mysqli或PDO代替mysql_*

Answer 2

值应使用引号。请试试这个;

$val='"' . implode('","',$name) . '"';
$sql= mysql_query("INSERT INTO incident_main(inc_patientName,inc_patientAge,inc_patientGender,inc_patientMobile,inc_patientAddress,inc_patientWard,inc_type,inc_diagnosis,inc_recurrence,inc_hospitalized,inc_treatment,inc_untraced)  VALUES($val);");

另外，在PHP 5.5中不推荐使用mysql_函数。使用PDO或mysqli代替mysql_函数。

http://php.net/manual/tr/function.mysql-query.php

Answer 3

要使用这些值，您可以在sql中将其替换为：

$sql = "INSERT INTO incident(inc_patientName,inc_patientAge,inc_patientGender,inc_patientMobile,inc_patientAddress,inc_patientWard,inc_type,inc_diagnosis,inc_recurrence,inc_hospitalized,inc_treatment,inc_untraced)". " VALUES( '".$val."')";

上面的代码将给出以下结果：

INSERT INTO 事件（inc_patientName，inc_patientAge，inc_patientGender，inc_patientMobile，inc_patientAddress，inc_patientWard，inc_type，inc_diagnosis，inc_recurrence，inc_hospitalized，inc_treatment，inc_untraced）VALUES（＆＃39; epoch＆＃39;，＆＃39; abc＆＃39;，＆＃39; 20＆＃39;，＆＃39;男＆＃39;，＆＃39;某些地址＆＃39;，＆＃39;字-1＆＃39;，＆＃39;某些数＆＃39;＆＃39;其他＆＃39）

但是，在您的代码中，您使用＆＃34;，＆＃34;来破坏值。，但在sql中字符串值用双引号＆＃34;＆＃34;
编码

要解决此问题，请按以下步骤更换代码：

$val=implode("','",$name);

将JSON解码值插入表中

3 个答案: