无法从Key JSON MySQL PHP获取值

Question

我正在使用PHP从MySQL数据库中检索数据，并尝试使用JSON.stringify和JSON.parse来创建对象。它工作正常，但我无法获得指定的键/值。只是整个对象。我把它分成几部分。这是PHP代码：

第一个 PHP 文件：

<?php
session_start();
include("web_db_operations.php");
if(isset($_POST['recipeId']) && isset($_SESSION['email'])){
    $recipeId = $_POST['recipeId'];
    $email = $_SESSION['email'];
}
else{
    echo "Did not work";
}
    $results = getAllMyRecipesAsList_recipeTable2($email, $recipeId);
    $_SESSION['recipeResults'] = $results;
    header('location:web_selected_recipe.php');
    exit();
?>

第二个 PHP 文件

 function getAllMyRecipesAsList_recipeTable2(string $email, int $recipeId){
    include 'config.php';
    $sql = 'SELECT * FROM recipeTable WHERE email = :email AND recipeId = :recipeId';
    $stmt = $conn->prepare($sql);       
    $stmt->bindParam(':email', $email, PDO::PARAM_STR);
    $stmt->bindParam(':recipeId', $recipeId, PDO::PARAM_STR);
    $stmt->execute();
    $getResults = $stmt->fetchAll(PDO::FETCH_ASSOC);
    $json = array();
    if(count($getResults) > 0){
        foreach($getResults as $row){
            $json[] = array('firstName' => $row['firstName'],
                           'lastName' => $row['lastName'],
                           'email' => $row['email'],
                           'recipeName' => $row['recipeName'],
                           'description' => $row['description'],
                           'ingredients' => $row['ingredients'],
                           'prepTime' => $row['prepTime'],
                           'steps' => $row['steps'],
                           'nutrition' => $row['nutrition'],
                           'servings' => $row['servings'],
                           'rating' => $row['rating'],
                           'tags' => $row['tags'],
                           'imagePath' => $row['imagePath'],
                           'imageName' => $row['imageName'],
                           'recipeId' => $row['recipeId']
                           );
         }
        $results = json_encode($json);
        return $results;
    }else{
        echo "no data found";
    }
    }

然后在我的JS文件中检索(这只是相关部分(：

<script>
    <?php $results = $_SESSION['recipeResults'];
var results = <?php echo $results; ?>;
    var toString = JSON.stringify(results);
    console.log(toString);
    var parsed = JSON.parse(toString);
    console.log(parsed);
</script>

记录结果为字符串生成以下内容：

[{"firstName":"Marcus","lastName":"Holden","email":"[e-mail]","recipeName":"Aloo Paratha","description":"","ingredients":"","prepTime":"25 Minutes","steps":"","nutrition":"","servings":"","rating":"","tags":"","imagePath":"../userRecipeImages","imageName":"9110164.jpg","recipeId":"1"}]

解析的日志记录会产生以下内容：

[{…}]
0:description:
"No Description", email "[e-mail]", firstName:"Marcus", imageName:"9110164.jpg", imagePath:"../userRecipeImages", ingredients:"Some Ingredients",lastName:"Holden", nutrition:"Not given", prepTime:"25 Minutes", rating:"5/10", recipeId:"1", recipeName:"Aloo Paratha", servings: "6", steps:"Your Steps Here", tags:"It's bread"

现在，我已经尝试了所有步骤来获取与键关联的值......例如，我这里的对象称为解析...所以我尝试解析.firstName..返回未定义...以及 Object.keys(已解析(。我似乎拿不到钥匙。我想像数组一样使用它...设置内容如下：

element.innerHTML = parsed[2]...等。

我在这里错过了什么？

Answer 1

我认为你做的比你需要的要多得多。数据以 JSON 编码对象的形式发送给您。只需使用它。

<script>
var results = <?php echo $_SESSION['recipeResults']; ?>;
var first_results = results[0]; // Each array member is one object from your result set
console.log( first_results.firstName );
console.log( results[0].firstName ); // OR specify which array index to interact directly
</script>