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javascript - 当详细信息传递给模态时,id undefined

转载 作者:行者123 更新时间:2023-11-29 20:54:08 24 4
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我有一个由特定用户创建的食谱表,当点击表中每一行的铅笔标记时,会显示一个模式,显示该特定食谱的详细信息,它应该允许用户编辑配方并将更新版本保存到数据库中。然而,尽管详细信息已正确传递给模态,但配方 id 似乎并未传递给模态,因为我试图将配方 id 输出到控制台,但它说配方 id 未定义。我试图调试此错误但无济于事。任何人都可以深入了解为什么会这样吗?

 //Recipe.js    

$('.editThis').on('click', function() {
var recipe_id = $(this).attr('data-id');
var request = $.ajax({
url: "ajax/displayRecipe.php",
type: "post",
dataType: 'json',
data: {recipe_id : recipe_id}
});

request.done(function (response, textStatus, jqXHR){
console.log("response " + JSON.stringify(response));
$('#name').val(response.name);
$('#date').val(response.date);
});

});


$('#editRecipe').click(function() {
var recipe_id = $(this).attr('data-id');
var name_input = $('#name').val();
var date_input = $('#date').val();

var request = $.ajax({
url: "ajax/updateRecipe.php",
type: "post",
data: {name : name_input, date : date_input, recipe_id : recipe_id},
dataType: 'json'
});

request.done(function (response, textStatus, jqXHR){
console.log(response);

});
});

//Recipe.php

<?php

$recipeObject = new recipeList($database); //Lets pass through our DB connection
$recipe = $recipeObject->getUserRecipes($_SESSION['userData']['user_id']);

foreach ($recipe as $key => $recipes) {
echo '<tr><td>'. $value['name'].'</td><td>'. $value['date'].'</td><td>'.'<a data-id = '.$value['recip_id'].' data-toggle="modal" class="edit editThis" data-target="#editRecipe"><i class="fa fa-pencil"></i></a>'.'</td></tr>';
}
?>

// editRecipe Modal

<div id="recipe" class="modal">
<div class="modal-dialog">
<div class="modal-content">
<div class="modal-header recipe">
<h1 class="modal-title">Edit Recipe</h4>
</div>
<div class="modal-body">
<form method="post" id="updateRecipeForm">

<?php
require_once('classes/recipes.classes.php');
$recipeObject = new recipeList($database);
$recipe = $recipeObject->getRecipeDetails(recipe_id);
if(isset($_POST['submit'])) {
$updateRecipe = $recipeObject ->updateRecipe($_POST['name'], $_POST['date'], $_POST['recipe_id']);
if($updateRecipe) {
echo ("Your recipe has been updated!";
}
}
?>

<div class="form-group">
<input type="text" class="control-form" id="name" value = "<?php echo $recipe['name']; ?>">
</div>
<div class="form-group">
<input type="date" class="control-form" id="date" value = "<?php echo $recipe['date']; ?>">
</div>
</div>
<div class="form-group">
<input type="hidden" class="form-control" data-id=".$recipe['recipe_id']." id="recipe_id" name="recipe_id" value = "<?php echo $recipe['recipe_id']; ?>">
</div>
<button type="submit" class="btn recipe" id="editRecipe" data-dismiss="modal">Save</button>
</form>
</div>
</div>
</div>
</div>


//ajax - updateRecipe.php



<?php
require_once('../includes/database.php');
require_once('../classes/recipes.classes.php');

if($_POST['name'] && $_POST['date'] && $_POST['trans_id']){
$recipeObject = new recipeList($database);
echo $recipeObject->updateRecipe($_POST['name'], $_POST['date'], $_POST['recipe_id']);
}
?>

//recipes.classes.php
...

public function getRecipeDetails($recipeid){
$query = "SELECT * FROM recipe WHERE recipe_id = :recipe_id";
$pdo = $this->db->prepare($query);
$pdo->bindParam(':recipe_id', $recipeid);
$pdo->execute();
return $pdo->fetch(PDO::FETCH_ASSOC);
}

public function updateRecipe($name, $date, $recipe_id){
$query = "UPDATE recipe SET name = :name, date = :date WHERE recipe_id = :recipe_id";
$pdo = $this->db->prepare($query);
$pdo->bindParam(':name', $name);
$pdo->bindParam(':date', $date);
$pdo->bindParam(':recipe_id', $recipe_id);
$pdo->execute();
}

最佳答案

尝试以下操作:

$(document).on('click', '.editThis',function() {...});
$(document).on('click','#editRecipe',function() {...});

关于javascript - 当详细信息传递给模态时,id undefined,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/50227646/

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