Saving JSON string to MySQL database

社会主义新天地 提交于 2019-11-30 04:03:40

We'll use json_decode json_decode documentation

Also be sure to escape! here's how I would do it below...

/* create a connection */
$mysqli = new mysqli("localhost", "root", null, "yourDatabase");

/* check connection */
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

/* let's say we're grabbing this from an HTTP GET or HTTP POST variable called jsonGiven... */
$jsonString = $_REQUEST['jsonGiven'];
/* but for the sake of an example let's just set the string here */
$jsonString = '{"name":"jack","school":"colorado state","city":"NJ","id":null}
';

/* use json_decode to create an array from json */
$jsonArray = json_decode($jsonString, true);

/* create a prepared statement */
if ($stmt = $mysqli->prepare('INSERT INTO test131 (name, school, city, id) VALUES (?,?,?,?)')) {

    /* bind parameters for markers */
    $stmt->bind_param("ssss", $jsonArray['name'], $jsonArray['school'], $jsonArray['city'], $jsonArray['id']);

    /* execute query */
    $stmt->execute();

    /* close statement */
    $stmt->close();
}

/* close connection */
$mysqli->close();

Hope this helps!

This is example for help you

<?php
 $json = '{"name":"jack","school":"colorado state","city":"NJ","id":null}';// You can get it from database,or Request parameter like $_GET,$_POST or $_REQUEST or something :p
 $json_array = json_decode($json);

 echo $json_array["name"];
 echo $json_array["school"];
 echo $json_array["city"];
 echo $json_array["id"];
?>

Hope this help !

Decode into an array and pass it in your mysql_query, the code below is not using mysql_real_escape_string or any other means of security, which you should implement.

Assume $json is {"name":"jack","school":"colorado state","city":"NJ","id":null}

$json_array = json_decode($json);

You now have indexes in a php array, such as: $json_array['name']

mysql_query("INSERT INTO student (name, school,city) VALUES('".$json_array['name']."', '".$json_array['school']."', '".$json_array['city']."') ") or die(mysql_error());  
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