Question

早安。我的问题是在点击更新按钮后，编辑元素就像标题，日期，内容和图像一样，只有回显输出显示为“已成功更新”。我想要做的是点击更新按钮后，元素将保持在那里，它将回应那里。

这是我的edit2.php

单击更新按钮后，只显示输出“Successfully update”，编辑元素消失

这是edit2.php的php代码

<?php

include_once('connection.php');

 $newsid = $_GET['news_id'];

    if(isset($_POST['esubmit'])){
        /* create a prepared statement */
        if ($stmt = mysqli_prepare($con, "SELECT * FROM news WHERE news_id = ? LIMIT 1")) {
            /* bind parameters */
            mysqli_stmt_bind_param($stmt, "s", $newsid);

            /* execute query */
            mysqli_stmt_execute($stmt);

            /* get the result set */
            $result = mysqli_stmt_get_result($stmt);

            /* fetch row from the result set */
            $row = mysqli_fetch_array($result);
        }

    }


    if(isset($_POST['update'])){


        if($_FILES['image']['error'] == 0) {
          $image= addslashes(file_get_contents($_FILES['image']['tmp_name']));
          $image_name = addslashes($_FILES['image']['name']);
          move_uploaded_file($_FILES["image"]["tmp_name"],"img/" . $_FILES["image"]["name"]);
          $newsimage="img/" . $_FILES["image"]["name"];

          $title = $_POST['titles'];
          $date = $_POST['dates'];
          $content = $_POST['contents'];

          $sql ="UPDATE news SET news_title ='$title', news_date ='$date', news_content = '$content', news_image ='$newsimage' WHERE news_id = '$newsid'";
          mysqli_query($con, $sql);
          echo "successfully updated";
        }

        else{
          $title = $_POST['titles'];
          $date = $_POST['dates'];
          $content = $_POST['contents'];
          $sql ="UPDATE news SET news_title ='$title', news_date ='$date', news_content = '$content' WHERE news_id = '$newsid'";
          mysqli_query($con, $sql);
          echo "successfully updated";
        }

    }
?>
<!DOCTYPE HTML>
<html>
<head>
</head>
<body>

<?php

    if(isset($_POST['esubmit'])){
        ?>

        <form method="post" enctype="multipart/form-data" action ="edit2.php?news_id=<?php echo $row['news_id']; ?>" >
            Title<input type ="text" name ="titles" value="<?php echo $row['news_title']; ?>"/><br>
            Date<input type ="text" name="dates" value="<?php echo $row['news_date']; ?>" /><br>
            Content<textarea name="contents"><?php echo $row['news_content']; ?></textarea>
            <input class="form-control" id="image" name="image" type="file" accept="image/*" onchange='AlertFilesize();'/>
            <img id="blah" src="<?php echo $row['news_image']; ?>" alt="your image" style="width:200px; height:140px;"/>

            <input type="submit" name="update" value="Update" />
        </form>

        <?php
    }

?>

<script src="js/jquery-1.12.4.min.js"></script>
<script src="js/bootstrap.min.js"></script>
<script type="text/javascript">
    function readURL(input) {
        if (input.files && input.files[0]) {
            var reader = new FileReader();

            reader.onload = function (e) {
                $('#blah').attr('src', e.target.result);
            }

            reader.readAsDataURL(input.files[0]);
        }
    }

    $("#image").change(function(){
        readURL(this);
    });
    </script>
</body>
</html>

Answer 1

问题在于这个if块，

if(isset($_POST['esubmit'])){ ...

当您提交表单时，$_POST['esubmit']将不会被设置，因此表单不会再次显示。所以你的if块应该是这样的：

if(isset($_POST['esubmit']) || isset($_POST['update'])){ ...

总的来说，您需要按以下方式更改第一个和第三个if块，

if(isset($_POST['esubmit'])){
    /* create a prepared statement */
    if ($stmt = mysqli_prepare($con, "SELECT * FROM news WHERE news_id = ? LIMIT 1")) {
        /* bind parameters */
        mysqli_stmt_bind_param($stmt, "s", $newsid);

        /* execute query */
        mysqli_stmt_execute($stmt);

        /* get the result set */
        $result = mysqli_stmt_get_result($stmt);

        /* fetch row from the result set */
        $row = mysqli_fetch_array($result);

        /* get all values */
        $title = $row['news_title'];
        $date = $row['news_date'];;
        $content = $row['news_content'];
        $newsimage = $row['news_image'];
    }

}

并且

if(isset($_POST['esubmit']) || isset($_POST['update'])){
    ?>

        <form method="post" enctype="multipart/form-data" action ="edit2.php?news_id=<?php echo $newsid; ?>" >
            Title<input type ="text" name ="titles" value="<?php if(isset($title)){ echo $title; } ?>"/><br>
            Date<input type ="text" name="dates" value="<?php if(isset($date)){ echo $date; } ?>" /><br>
            Content<textarea name="contents"><?php if(isset($content)){ echo $content; } ?></textarea>
            <input class="form-control" id="image" name="image" type="file" accept="image/*" onchange='AlertFilesize();'/>
            <img id="blah" src="<?php if(isset($newsimage)){ echo $newsimage; } ?>" alt="your image" style="width:200px; height:140px;"/>

            <input type="submit" name="update" value="Update" />
        </form>

    <?php
}

从扩展讨论中

由于您可以将图片上传设为可选内容，因此您需要在处理表单的else块中获取图片详细信息，以防用户未上传图片，如下所示：

if(isset($_POST['update'])){
    if($_FILES['image']['error'] == 0) {

        // your code

    }else{
        // your code

        /* get the image details*/
        if ($stmt = mysqli_prepare($con, "SELECT news_image FROM news WHERE news_id = ? LIMIT 1")) {
            mysqli_stmt_bind_param($stmt, "s", $newsid);
            mysqli_stmt_execute($stmt);
            $result = mysqli_stmt_get_result($stmt);
            $row = mysqli_fetch_array($result);
            $newsimage = $row['news_image'];
        }
    }
}

PHP - 更新后编辑元素消失了

1 个答案: