使用 JSON 将记录插入数据库

Question

以下是我的代码

<?php
$json = '{"apples":"green","bananas":"yellow"}';
$var=(json_decode($json, true));
print_r($var);

$username = "root";
$password = "";
$hostname = "localhost"; 

//connection to the eventbase
$dbhandle = mysql_connect($hostname, $username, $password) 
  or die("Unable to connect to MySQL");
echo "Connected to MySQL<br>";

//select a eventbase to work with
$selected = mysql_select_db("json",$dbhandle) 
  or die("Could not select json");

// Insert $event array into eventbase
    foreach ($json as $key => $value) {
    $db_insert = mysql_query("INSERT INTO fruits (fruit,color) VALUES" . $value);
    mysql_query($db_insert);
    if (!$db_insert)
    {
    die('Could not connect - event insert failed: ' . mysql_error());
    }
    }
?>

JSON 和 PHP 新手，请发布我可以做的任何更改以将这些记录插入 MySQL。

但是，另一部分仍然存在。我想将 JSON 编码的数据以字符串格式保存到 MySQL 中。我该怎么做？

Answer 1

如下使用$var，

foreach ($var as $fruit => $color) {
    $db_insert = mysql_query("INSERT INTO fruits (fruit,color) VALUES ('$fruit','$color')");
    .....

注意： Please, don't use mysql_* functions in new code。它们不再维护and are officially deprecated。看到red box？改为了解 prepared statements，并使用 PDO 或 MySQLi - this article 将帮助您决定哪个。如果你选择 PDO，here is a good tutorial。

Answer 2

我认为您没有在 for 循环中使用正确的变量。在 json 解码之后，您在 for 循环中使用了相同的变量。 （你应该使用 $var 而不是 $json）我猜。

<?php
$json = '{"apples":"green","bananas":"yellow"}';
$var=(json_decode($json, true));
//print_r($var);

$username = "root";
$password = "";
$hostname = "localhost"; 

//connection to the eventbase
$dbhandle = mysql_connect($hostname, $username, $password) 
  or die("Unable to connect to MySQL");
echo "Connected to MySQL<br>";

//select a eventbase to work with
$selected = mysql_select_db("json",$dbhandle) 
  or die("Could not select json");

// Insert $event array into eventbase
    foreach ($var as $key => $value) {
    $db_insert = mysql_query("INSERT INTO fruits (fruit,color) VALUES" . $value);
    mysql_query($db_insert);
    if (!$db_insert)
    {
    die('Could not connect - event insert failed: ' . mysql_error());
    }
    }
?>