JSON 到 MYSQL - JSON 响应格式是否正确 - 正确循环？

Question

好的 - 我正在尝试加载 JSON 响应（来自名为recipes.json 的外部文件，其中包含数百个食谱），其格式如下，以便将其插入名为“recipes”的 MySQL 表中：

{ "recipeName": "After Glow Smoothie",  "ingredients": "4 oz. (1/2 cup) pomegranate juice", "ingredients2": "4 oz. (1/2 cup orange juice)", "ingredients3": "2 scoops Vi-Shape shake mix", "ingredients4": "1 cup frozen pineapple", "ingredients5": "5 ice cubes"},
{ "recipeName": "All Berry Delight",     "ingredients": "8 oz. skim milk", "ingredients2": "2 scoops Vi-Shape shake mix", "ingredients3": "1/4 cup frozen raspberries", "ingredients4": "1/4 cup frozen blackberries", "ingredients5": "1/4 cup frozen strawberries", "ingredients6": "1/4 cup frozen dark cherries", "ingredients7": "5 ice cubes"}

我不太方便使用数组，所以我想知道为什么我无法循环遍历食谱并正确插入它们。是我的 JSON 格式错误，还是我是一个 PHP 菜鸟，以至于我在主代码中犯了错误。参考如下：

<?php
header('Content-Type: text/html; charset=utf-8');

$hostname_ndb = "localhost";
$database_ndb = "test";
$username_ndb = "root";
$password_ndb = "root";
$ndb = mysql_pconnect($hostname_ndb, $username_ndb, $password_ndb) or trigger_error(mysql_error(),E_USER_ERROR); 

$url = "http://localhost:8888/shakerecipes/recipes.json";


$json = file_get_contents($url);
// var_dump(json_decode($json, true));
$out = json_decode($json, true);


foreach($out["recipeName"] as $recipeNames) { 
$name = addslashes($recipeNames[recipeName]); 
$ingredients= addslashes($recipeNames[ingredients]); 
$ingredients2 = addslashes($recipeNames[ingredients2]);
$ingredients3 = addslashes($recipeNames[ingredients3]);
$ingredients4 = addslashes($recipeNames[ingredients4]);
$ingredients5 = addslashes($recipeNames[ingredients5]);
$ingredients6 = addslashes($recipeNames[ingredients6]);
$ingredients7 = addslashes($recipeNames[ingredients7]);
$ingredients8 = addslashes($recipeNames[ingredients8]);
$ingredients9 = addslashes($recipeNames[ingredients9]);

mysql_query("INSERT INTO test (recipeName, ingredients, ingredients2, ingredients3, ingredients4, ingredients5, ingredients6, ingredients7, ingredients8, ingredients9) VALUES('$name', '$ingredients', '$ingredients2', '$ingredients3', '$ingredients4', '$ingredients5', '$ingredients6', '$ingredients7', '$ingredients8', '$ingredients9')") or die (mysql_error()); 
}
?>

感谢您的任何和所有提示/帮助。

BRR

Answer 1

首先，您应该使用mysql_real_escape_string而不是addslashes 。

其次，您应该/可以使用 $recipeNames 执行另一个 foreach 循环。

或者你可以使用 lambda/closure 风格。

array_walk($recipeNames, function(&$value) {
    $value = mysql_real_escape_string($value);
});

之后你的价值观就会崩溃

mysql_query("INSERT INTO test (recipeName, ingredients, ingredients2, ingredients3, ingredients4, ingredients5, ingredients6, ingredients7, ingredients8, ingredients9) VALUES('".implode('\',\'', $recipeNames)."')") or die (mysql_error());

Answer 2

对于那些想知道什么有效的人 - 这是我想出的：

<?php
header('Content-Type: text/html; charset=utf-8');

$hostname_ndb = "localhost";
$database_ndb = "test";
$username_ndb = "root";
$password_ndb = "root";
$ndb = mysql_pconnect($hostname_ndb, $username_ndb, $password_ndb) or trigger_error(mysql_error(),E_USER_ERROR); 
mysql_select_db($database_ndb);

$url = "http://localhost:8888/shakerecipes/recipes.json";
$json = file_get_contents($url);

$out = json_decode($json, true);

foreach($out["recipes"] as $recipe) { 
$name = addslashes($recipe[recipeName]); 
$ingredients= addslashes($recipe[ingredients]); 
$ingredients2 = addslashes($recipe[ingredients2]);
$ingredients3 = addslashes($recipe[ingredients3]);
$ingredients4 = addslashes($recipe[ingredients4]);
$ingredients5 = addslashes($recipe[ingredients5]);
$ingredients6 = addslashes($recipe[ingredients6]);
$ingredients7 = addslashes($recipe[ingredients7]);
$ingredients8 = addslashes($recipe[ingredients8]);
$ingredients9 = addslashes($recipe[ingredients9]);

mysql_query("INSERT INTO recipes (recipeName, ingredients, ingredients2, ingredients3, ingredients4, ingredients5, ingredients6, ingredients7, ingredients8, ingredients9) VALUES('$name', '$ingredients', '$ingredients2', '$ingredients3', '$ingredients4', '$ingredients5', '$ingredients6', '$ingredients7', '$ingredients8', '$ingredients9')") or die (mysql_error()); 

}
?>