[英]form doesn't contains a value to let me edit the file
嗨,大家好,我的表格存在問題,我已經從數據庫中設置了一個值,但文件輸入未從數據庫中顯示出來。 誰知道什么問題? 我在mysql中用於文件的數據類型是中等大小的Blob,它將文件存儲在名為upload的文件夾中。 第一個代碼是我的editquiz.php,而第二個代碼是我的pedit.php。
<form method ="post" action = "peditQuiz.php" enctype="multipart/form-data">
<input type = "hidden" name = "quizID" id="quizID" value = "<?php echo $st_row['q_id'] ?>" >
<div class="form-group">
<h4><b>Quiz ID: <span class="text-primary"><?php echo $st_row['q_id'] ?></span> </b></h4>
</div>
<hr>
<div class="form-group">
<label>Quiz Title</label>
<input type="text" class="form-control" name = "quizTitle" id="quizTitle" value = "<?php echo $st_row['q_title'] ?>" required>
</div>
<div class="form-group">
<label>Quiz Description</label>
<input type="text" class="form-control" name = "quizDesc" id="quizDesc" value = "<?php echo $st_row['q_desc'] ?>" required >
</div>
<div class="form-group">
<label>Quiz URL (paste the link here)</label>
<input type="url" class="form-control" name = "quizURL" id="quizURL" value = "<?php echo $st_row['q_url'] ?>">
</div>
<div class="form-group">
<label>Upload new Quiz file (Max. allowed file size is 8MB)</label>
<input type="file" class="form-control" name = "quizFile" id ="quizFile" value = "<?php echo $st_row['q_file'] ?>" placeholder = "<?php echo $st_row['q_file'] ?>">
</div>
<input type="submit" class="btn btn-default" name = "btnUpdate" value = "Update">
<input type="reset" class="btn btn-default" value = "Clear">
<a href = "manageQuiz.php"><button type="button" style = "float:right" class="btn btn-info" >Back</button></a>
//Pedit.php
<?php
include("connection.php");
$userid = $_SESSION['userID'];
$title= $_POST['quiz_Title'];
$desc = $_POST['quiz_Desc'];
$url = $_POST['quiz_URL'];
$file = rand(1000, 100000). "-".$_FILES['quiz_File']['name'];
$file_loc = $_FILES['quiz_File']['tmp_name'];
$file_size = $_FILES['quiz_File']['size'];
$file_type = $_FILES['quiz_File']['type'];
$folder="files/";
move_uploaded_file($file_loc, $folder.$file);
/*
$id = $_POST['quizID'];
$sql = "SELECT * FROM quiz where quiz_id = '$id'";
$result = mysql_query($sql) or die(mysql_error());
$row = mysql_fetch_assoc($result);
$count = mysql_num_rows($result);
if($count > 0){
echo "<script>alert('Quiz record already exist');window.location.href = 'addQuiz.php';</script>";
} else { */
if($url==NULL){
$sql = "insert into quiz (q_title, q_desc, q_url, q_file, admin)
values ('$title','$desc ','$url','$file','$userid ' )" ;
mysql_query($sql);
echo "<script>alert('New record created succcessfully');window.location.href = 'manageQuiz.php';</script>";
} else{
$sql = "insert into quiz (q_title, q_desc, q_url, admin)
values ('$title','$desc ','$url','$userid ' )" ;
mysql_query($sql);
echo "<script>alert('New record created succcessfully');window.location.href = 'manageQuiz.php';</script>";
}
//}
mysql_close($con);
?>
您的問題很難理解,但我會嘗試:
看起來您正在使用php通過PHP轉儲表單中的值,然后再通過include語句加載值。
我也不確定為什么要使用基於文件的數據庫,但似乎也包含sql命令,但是無論如何將值加載到“ $ st_row ['q_id']”中,都必須先加載它們嘗試將它們回顯到您的html中。
如果出於某種原因以后需要包含db文件,則可以使用javascript將值推到事實之后的表單字段中。
但是,如果您要從包含的文件中查找sql查詢的結果...我想您需要指定期望從哪個文件中加載什么值,並提供該代碼。
希望能有所幫助。 祝好運。 還恭喜您問了關於stackoverflow的問題。 看起來您是個初學者,但正在努力。 ;)
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.