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請幫助任何人。我一直在調試幾天。我只是有一個問題,我會很高興,如果你幫助我。我的圖片沒有上傳到我的數據庫
我的圖像無法上傳到我的數據庫。它甚至不會給我一個錯誤。它只是上傳我的查詢的其他組件,將圖像留在後面。這裏是我的代碼
<?php
global $connection;
if(isset($_POST['submit'])){
$post_category_id = $_POST['post_category_id'];
$post_title = $_POST['post_title'];
$post_author = $_POST ['post_author'];
$post_status = $_POST['post_status'];
$post_image = isset($_FILES['post_image']['image_name']);
$post_image_temp = isset($_FILES['post_image']['temp_name']);
$post_tags = $_POST['post_tags'];
$post_comment_count = 4;
$post_date = date('d-m-y');
$post_content = $_POST['post_content'];
$post_status = $_POST['post_status'];
move_uploaded_file($post_image_temp, '../images/$post_image');
$query = "INSERT INTO posts (post_category_id, post_title, post_author, post_status, post_image, post_tags, post_comment_count, post_date, post_content) VALUES ('{$post_category_id}', '{$post_title}', '{$post_author}', '{$post_status}', '{$post_image}', '{$post_tags}', {$post_comment_count}, now(), '{$post_content}') ";
$result = mysqli_query($connection, $query);
if ($result){
echo "Post Published";
} else {
echo "(Error Code:" . $_FILES['post_image']['error'] . ")";
}
}
?>
<form action="" method="post" enctype="multipart/form-data">
<div class="form-group">
<label for="post_title">Title</label>
<input type="text" name ='post_title' class="form-control" >
</div>
<div class="form-group">
<label for="Post_author">Post Author</label>
<input type="text" name ='post_author' class="form-control">
</div>
<div class="form-group">
<label for="post_category_id">Post Category</label>
<input type="text" name ='post_category_id'class="form-control" >
</div>
<div class="form-group">
<label for="post_status">Post Status</label>
<input type="text" name ='post_status' class="form-control">
</div>
<div class="form-group">
<label for="post_image">Upload Image</label>
<input type="file" name ="post_image" id="post-image" class="form-control" >
</div>
<div class="post tags">
<label for="post_tags">Post Tags</label>
<input type="text" name ='post_tags'class="form-control" >
</div>
<div class="post comment count">
<label for="post_comment_count">Post Comment Count</label>
<input type="text" name ='post_comment_count'class="form-control" >
</div>
<div class="form-group">
<label for="post_date">post date</label>
<input type="date" name ='post_date'class="form-control" >
</div>
<div class="form-group">
<label for="post_content">Post content</label>
<textarea name="post_content" id="" cols="30" rows="10"></textarea>
</div>
<input class= "btn btn-primary"type="submit" name = "submit" value="Publish Post">
</form>
請注意我已經完全控制權限到我從上傳的位置。我相信你會發現我的IMAGE_NAME的isset功能和image_temp_name這是因爲沒有它,我只是得到一個
未定義的變量錯誤
點擊我的提交按鈕後
瞭解準備好的語句以防止sql注入 – Jens
您的腳本處於危險中[SQL注入攻擊](http://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php) 即使[如果你逃避輸入,它不安全!](http :http://stackoverflow.com/questions/5741187/sql-injection-that-gets-around-mysql-real-escape-string) 使用[編寫參數化語句](http://php.net/manual/en/mysqli .quickstart.prepared-statements.php) – RiggsFolly
isset不會賦值給變量。 –