Question

我有一个论坛，用户可以在该论坛中输入他们正在寻找的作业，然后将其提交到数据库，然后显示在下一页上。只有我无法上传任何数据，我不知道为什么。

我也在努力寻找错误检查的方法。有什么想法吗？

// Check for job submission
        if(isset($_POST['submit']))

            //empty error array
            $error = array();

        // check for a things in the field

        if(empty($_POST['job']))
        {
            $error[] = 'Please fill in all required fields';

        }

        // iff there are no errors, insert the job listing into the database. 
        // otherwies, display error. 
        if(sizeof($error) == 0)
        {
            // insert job listing
            $query = "INSERT INTO job (
                                job_id, 
                                user_id, 
                                jobtitle,
                                company, 
                                location, 
                                summary,
                                responsibilities, 
                                skills
                                ) VALUES (
                                  null, 
                                  '{$_SESSION['user_id']}',
                                   '{$_POST['jobtitle']}', 
                                   '{$_POST['company']}', 
                                   '{$_POST['location']}', 
                                   '{$_POST['summary']}', 
                                   '{$_POST['responsibilities']}',
                                   '{$_POST['skills']}',
                                    NOW()
                                    )";

            $result = mysqli_query($dbc, $query) or die('Query failed: ' . mysqli_error($dbc));

            // display a confirmation
            echo "<div class=\"alert alert success\">Your job listing has been added</div>";

        } else {

            // display error message
            foreach($error as $value)
            {
                echo "<div class=\"alert alert-danger\"{$value}</div>";
            }
        }

        ?>

        <!-- Job listing Form -->
        <form method="post" action="listings.php">
            <div class="form-group">
                <label> Job Title </label>
                <input name ="jobtitle" type="text" class="jobform"/>

                <label>Company/Organization</label>
                <input name="company" type="text" class="jobform"/>

                <label> Job Location </label>
                <input name ="location" type="text" class="jobform"/>

                <label> Job Responsibilities </label>
                <textarea name="summary" rows="8" cols="20" class="jobfourm"></textarea>

                <label> Job Responsibilities </label>
                <textarea name="responsibilities" rows="8" cols="20" class="jobfourm"></textarea>

                <label> Job Skills </label>
                <textarea name="skills" rows="8" cols="20" class="jobforum"></textarea>

            </div>

        <div class="form-group">
                    <input name="submit" type="submit" value="Submit" class="btn btn-large btn-primary" />
                </div>

        </form>

     </div>

Answer 1

我的下注是你的问题：

 (
                                job_id, 
                                user_id, 
                                jobtitle,
                                company, 
                                location, 
                                summary,
                                responsibilities, 
                                skills
                                ) VALUES (
                                  null, 
                                  '{$_SESSION['user_id']}',
                                   '{$_POST['jobtitle']}', 
                                   '{$_POST['company']}', 
                                   '{$_POST['location']}', 
                                   '{$_POST['summary']}', 
                                   '{$_POST['responsibilities']}',
                                   '{$_POST['skills']}',
                                    NOW()

对于什么应该是job_id，你传递null。现在，我将假设所有工作都必须有一个工作ID，对吗？你需要实际传入一个有效的id，因为我打算把钱（或代表）押在桌子上是一个不可为空的字段。此外，您已在值中添加了一列未在列名参数中声明的列。

作业Entrys不提交到数据库表

1 个答案: