JSON到MYSQL - JSON响应格式正确 - 循环正确吗?

时间:2011-05-10 01:04:25

标签: php mysql arrays json

好的 - 我正在尝试加载一个JSON响应(来自一个名为recipes.json的外部文件,其中包含数百个配方),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"}

我对数组不太方便所以我想知道为什么我无法循环配方并正确插入它们。我的JSON格式不正确,还是我这样的PHP菜鸟,我在主代码中犯了错误。供参考,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()); 
}
?>

感谢您提供的所有提示/帮助。

BRR

2 个答案:

答案 0 :(得分:1)

首先,您应该使用mysql_real_escape_string而不是addslashes。

其次你应该/可以用$ recipeNames预先形成另一个foreach循环。

或者你可以做lambda / closure风格。

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

之后你可以破坏你的价值观

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

答案 1 :(得分:0)

对于那些想知道什么有用的人 - 这就是我想出的:

<?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()); 

}
?>