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你好編程的怪物，美好的一天！我想要做的是，當用戶點擊文件按鈕並上傳圖片時，它會更新，否則，如果用戶沒有更改它只是回顯該圖片。但是，當用戶沒有改變圖像，即時通訊有錯誤**Warning**: file_get_contents(): Filename cannot be empty in C:\xampp\htdocs\studentportal\edit2.php on line 27有人可以幫助我嗎？ if(isset($_FILES['image']))正常工作，但else statement不正確。如果用戶沒有改變它，我怎麼才能回顯那個圖片？即時通訊新的PHP，並開始學習它，請給我的想法。PHP - 不能使用抓取的數據
這是在else語句中的錯誤原因$newsimages = $row['news_image'];。
else{
$title = $_POST['titles'];
$date = $_POST['dates'];
$content = $_POST['contents'];
$newsimages = $row['news_image'];
$sql ="UPDATE news SET news_title ='$title', news_date ='$date', news_content = '$content', news_image ='$newsimages' WHERE news_id = '$newsid'";
mysqli_query($con, $sql);
echo "oh it worked again ";
}
這是所有的PHP代碼
<?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($_POST['update'])){
if(isset($_FILES['image'])){
$file=$_FILES['image']['tmp_name'];
$image= addslashes(file_get_contents($_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 "oh it worked ";
}
else{
$title = $_POST['titles'];
$date = $_POST['dates'];
$content = $_POST['contents'];
$newsimages = $row['news_image'];
$sql ="UPDATE news SET news_title ='$title', news_date ='$date', news_content = '$content', news_image ='$newsimages' WHERE news_id = '$newsid'";
mysqli_query($con, $sql);
echo "oh it worked again ";
}
}
?>
<!DOCTYPE HTML>
<html>
<head>
</head>
<body>
<?php
if(isset($_POST['esubmit'])){
?>
<form method="post" action ="edit2.php?news_id=<?php echo $row['news_id']; ?>" enctype="multipart/form-data">
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]);
}
}
$("#image").change(function(){
readURL(this);
});
</script>
</body>
</html>
你不想檢查'isset（$ _ FILES ['image']）'因爲默認情況下當你提交表單時它將被設置。您可能希望檢查數組$ _FILES ['image'] ['error']'中的值的狀態。 – Rasclatt
也不要這樣做，''更新新聞SET news_title ='$ title'，news_date ='$ date'，news_content ='$ content'，news_image ='$ newsimage'WHERE news_id ='$ newsid'「'it很容易發生SQL注入。任何包含在'$ _POST'中的字符都可能會讓你的sql語句變得糟糕。處理用戶提交的輸入時使用綁定參數。 – Rasclatt
注意，如果你在'if'和'else'中都有相同的行，只需刪除一個集合，並將其他集合移出'if'和'else'，因爲事件發生在兩種情況下。 'if'和'else'中只有不同的/改變的元素。保存幾行。 – Rasclatt