我的代码在这里有什么问题吗？ [重复]

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• The 3 different equals 5 answers
• unexpected error message in php form (SQL syntax error) 3 answers

whenever I want to insert data with my form ,everything woks fine except the content field that does not insert anything to db!

Here's my full code:

    <?php 
if (isset($_POST['submit'])){
    $post_title = $_POST['title'];
    $post_date = date('d-m-y');
    $post_author = $_POST['author'];
    $post_keywords = $_POST['keywords'];
    $post_content = $_POST['content'];
    $post_image = $_FILES['image']['name'];
    $image_tmp = $_FILES['image']['tmp_name'];
    $post_summary = $_POST['summary'];

    if ($post_title=='' || $post_keywords=='' || $post_content='' || $post_author=='' || $post_summary==''){
        echo '<script>alert("Some fields are missing")</script>';
    }else{
        move_uploaded_file($image_tmp,"post_images/$post_image");
        $insert_query = "INSERT INTO posts 
    (post_title, post_date, post_author, post_image, post_keywords, post_content, post_summary)
    VALUES ('$post_title', '$post_date', '$post_author', '$post_image', '$post_keywords', '$post_content' , '$post_summary')";
        $insert_post = mysqli_query($con,$insert_query);
        if ($insert_post){
            echo '<h3 style="color:green">Post has been added successfully.</h3>';
        }elseif (!$insert_post){
            echo mysqli_error($con);
        }else{
            echo "Somthing goes wrong! Please contact with our support team...";
        }
    }
}
?>
<form method="POST" action="" enctype="multipart/form-data">
    <table width="600" align="center" border="10">
        <tr>
            <td align="center"><h6>Insert Post Title</h6></td>
            <td align="center"><input type="text" name="title"/></td></br>
        </tr>
        <tr>
            <td align="center"><h6>Insert Post Author</h6></td>
            <td align="center"><input type="text" name="author"/></td></br>
        </tr>
        <tr>
            <td align="center"><h6>Insert Post Keywords</h6></td>
            <td align="center"><input type="text" name="keywords"/></td></br>
        </tr>
        <tr>
            <td align="center"><h6>Insert Post Image</h6></td>
            <td align="center"><input type="file" name="image"/></td></br>
        </tr>
        <tr>
            <td align="center"><h6>Insert Post Content</h6></td>
            <td align="center"><textarea name="content" cols="10" rows="10"></textarea></td></br>
        </tr>
        <tr>
            <td align="center"><h6>Insert Post Summary</h6></td>
            <td align="center"><textarea name="summary" cols="5" rows="5"></textarea></td></br>
        </tr>
        <tr>
            <td align="center"><input type="submit" name="submit" value="Submit"/></td>
        </tr>
    </table>
</form>

The problem is that the $post_content variable does not insert anything into post_content field in my table however the other variables works correctly & also the success message (line 21) appears.

Here's my table structure:

post_id => int(11)

post_title => varchar(100)

post_date => date

post_author => varchar(100)

post_image => image

post_keywords => text

post_content => text

post_summary => text

</div>

Answer 1

The problem is that you have specified the assignment instead of comparison in the following line:

if ($post_title=='' || $post_keywords=='' || $post_content='' || $post_author=='' || $post_summary==''){

Change $post_content='' to $post_content == ''