Question

在我的php中，我正在运行一个简单的查询，它从我拥有的数据库中返回一个结果集（0或很多）。

目前正面朝下，rersult看起来像这样：

name: Smoothie description: Banana Smothie name: Phad Thai description: Noodles with shrimps name: Noodles description: Noodles with noodles.

字符串也可以看起来像这样，name: Smoothie description: Banana Smothie或更多条目，如上例所示。

下面的代码告诉我：

[{"name":"Smoothie","description":"Banana Smothie"}][{"name":"Phad Thai","description":"Noodles with shrimps"}]

我想拥有的是，所以它可以只是一个json对象：

 [{"name":"Smoothie","description":"Banana Smothie"},{"name":"Phad Thai","description":"Noodles with shrimps"}]

这是我的php：

    <?php
 include_once 'db/dbconnect.php';
 $input = json_decode(stripcslashes($_POST['data']));

 for ($i=0; $i < count($input); $i++) {
     $stmt=$con->prepare("SELECT recipes.recipeName, recipes.recipeDescription FROM ingredients, recipes, recipesingredients WHERE recipes.recipeId = recipesingredients.recipeIdFK AND recipesingredients.ingredientIdFK = ingredients.IngredientId AND ingredients.ingredientName = ?");
     $stmt->bind_param("s", $input[$i]);
     $stmt->execute();
     $stmt->store_result();
     $stmt->bind_result($db_recipe_name, $db_recipe_description);

     $rslt = array();
     $arr = 0;
     while ($stmt->fetch()) {
         $rslt[$arr] = array('name' => $db_recipe_name, 'description' => $db_recipe_description);
         $arr++;
     }
     $jsonRslt = json_encode($rslt);
     echo $jsonRslt;
 }
 ?>

有人可以帮我把它变成一个json对象吗？

Answer 1

不是在循环内创建json_encode和echo的新数组，而是在循环之前创建一个并将每个对象添加到循环中。

示例：

<?php
include_once 'db/dbconnect.php';
$input = json_decode(stripcslashes($_POST['data']));

// Create an array before the loop
$json = [];

for ($i=0; $i < count($input); $i++) {
    $stmt=$con->prepare("SELECT recipes.recipeName, recipes.recipeDescription FROM ingredients, recipes, recipesingredients WHERE recipes.recipeId = recipesingredients.recipeIdFK AND recipesingredients.ingredientIdFK = ingredients.IngredientId AND ingredients.ingredientName = ?");
    $stmt->bind_param("s", $input[$i]);
    $stmt->execute();
    $stmt->store_result();
    $stmt->bind_result($db_recipe_name, $db_recipe_description);

    while ($stmt->fetch()) {
        // Add each element to the main array
        $json[] = array('name' => $db_recipe_name, 'description' => $db_recipe_description);
    }
}

// json_encode and echo the main array
echo json_encode($json);
?>

Answer 2

在for循环内，每次为$rslt重新初始化i数组，这就是为什么要获取多个JSON对象而不是单个JSON对象的原因。

您需要在$rslt循环之外初始化for并在for循环后将其编码为JSON。

<?php
 include_once 'db/dbconnect.php';
 $input = json_decode(stripcslashes($_POST['data']));

 // Initialize array here
 $rslt = array();
 $arr = 0;

 for ($i=0; $i < count($input); $i++) {
     $stmt=$con->prepare("SELECT recipes.recipeName, recipes.recipeDescription FROM ingredients, recipes, recipesingredients WHERE recipes.recipeId = recipesingredients.recipeIdFK AND recipesingredients.ingredientIdFK = ingredients.IngredientId AND ingredients.ingredientName = ?");
     $stmt->bind_param("s", $input[$i]);
     $stmt->execute();
     $stmt->store_result();
     $stmt->bind_result($db_recipe_name, $db_recipe_description);

     while ($stmt->fetch()) {
         $rslt[$arr] = array('name' => $db_recipe_name, 'description' => $db_recipe_description);
         $arr++;
     }
 }

 // encode into JSON
 $jsonRslt = json_encode($rslt);
 echo $jsonRslt;
 ?>

使用php将db结果集转换为有效的json

2 个答案: