将一个数据库结果集转换为一个有效的json和php

Question

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

目前在门前的rersult看起来是这样的：

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

的字符串也可以是这样的，又名name: Smoothie description: Banana Smothie或多个条目，就像上面的例子。

下面的代码给了我这样的：

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

我想有是，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; 
} 
?> 

有人可以帮我把它做成一个json对象吗？

Answer 1

除了在循环中创建json_encode和echo新的数组之外，还可以在循环之前创建一个数组，并将每个对象添加到循环中。

实施例：

<?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循环中创建php数组。不要json在循环内编码/回显。回声只是最终产生的PHP数组外侧for循环。另外，将响应头设置为json。 - 抱歉打字错误。从手机回答。

Answer 3

里面for循环中，$rslt阵列被重新初始化为每个i每次这就是为什么你得到多个JSON对象而不是单个的。

您需要初始化for循环之外的$rslt，并在for循环之后将其编码为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; 
?>