大家好,
我有一个由10列组成的表,其中4列我想插入图像路径。
任何人都可以指导我如何将图像路径与其他6个列数据一起插入数据库,并将图像上传到服务器(文件夹内)。
这里是我的HTML代码:
<div class="panel-body">
<div class="form-group">
<label style="align-content:center" for="inputdefault">Product full name</label>
<input class="form-control" id="inputdefault" name="name2" type="text">
</div>
<div class="panel-body">
<div class="form-group">
<label style="align-content:center" for="inputdefault">Product category</label>
<input class="form-control" id="inputdefault2" name="name1" type="text">
</div>
<div class="panel-body">
<div class="form-group">
<label style="align-content:center" for="inputdefault">Product qty</label>
<input class="form-control" id="inputdefault3" name="name3" type="text">
</div>
<div class="panel-body">
<div class="form-group">
<label style="align-content:center" for="inputdefault">Picture 1.</label>
<input type="file" id="file3" name="files[]" multiple="multiple" accept="image/*" />
</div>
<div class="form-group">
<label style="align-content:center" for="inputdefault">Picture 2.</label>
<input type="file" id="file3" name="files[]" multiple="multiple" accept="image/*" />
</div>
<div class="form-group">
<label style="align-content:center" for="inputdefault">Picture 3.</label>
<input type="file" id="file3" name="files[]" multiple="multiple" accept="image/*" />
</div>
<div class="form-group">
<label style="align-content:center" for="inputdefault">Picture 4.</label>
<input type="file" id="file3" name="files[]" multiple="multiple" accept="image/*" />
</div>
这里是php代码: 注意:我没有将任何变量中的图像文件名添加到查询中,一切正常,只需要处理图像上传并将路径存储到数据库中
<?php
if(isset($_POST) and $_SERVER['REQUEST_METHOD'] == "POST")
{
$pname= stripslashes($_REQUEST['name1']);
$pcat= stripslashes($_REQUEST['name2']);
$pprice= stripslashes($_REQUEST['name3']);
$pqty= stripslashes($_REQUEST['qty']);
$pdesc=stripslashes($_REQUEST['description']);
//$spassword=stripslashes($_REQUEST['img']);
$sqlinsert ="INSERT INTO `noorizone`.`products` VALUES ('$pname', '$pcat', '$pprice')";
if($con-> query($sqlinsert)=== true)
{
echo "<center><b style='color:green;'> added successfully... </b>";
echo"</center>";
}
else{
echo "<b style='color:red;'> cant register". $con->error."</b>";
}
}
?>
感谢提前!!!
答案 0 :(得分:0)
这是我的例子,你可以学习我的代码:
$fileImageStatus = ""; // Create your variable for files
if(isset($_FILES['files_status']['name'])){ // Checking value for all input
$fileImageName = array(); // Make an array
for($i = 0; $i < count($_FILES['files_status']['name']); $i++){ // Looping
$imageName = $_FILES['files_status']['name'][$i]; // Give a variable with index from array looped
$uploadPath = '././sistem/users/members/'.$idPenerima.'/unggahan/';
move_uploaded_file($imageTmp, $uploadPath.$imageName); // Path upload
$fileImageName[] = $imageName; // Setting a values from array
}
$fileImageStatus = implode(",", $fileImageName); // Make that values an string
}
// in here you can insert to database with value from $fileImageStatus
答案 1 :(得分:0)
我认为如果您将文件名更改为file1而不是file [],那么您将没事。
中的示例
<!DOCTYPE html>
<html>
<body>
<!--
<?php
$target_dir = "uploads/";
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
$uploadOk = 1;
$imageFileType = pathinfo($target_file,PATHINFO_EXTENSION);
// Check if image file is a actual image or fake image
if(isset($_POST["submit"])) {
$check = getimagesize($_FILES["fileToUpload"]["tmp_name"]);
if($check !== false) {
echo "File is an image - " . $check["mime"] . ".";
$uploadOk = 1;
} else {
echo "File is not an image.";
$uploadOk = 0;
}
}
?>
-->
<form action="upload.php" method="post" enctype="multipart/form-data">
Select image to upload:
<input type="file" name="fileToUpload" id="fileToUpload">
<input type="submit" value="Upload Image" name="submit">
</form>
</body>
</html>
答案 2 :(得分:-1)
这是一个关于如何将多个图像上传到PHP和Mysql的好教程
http://www.wdb24.com/how-to-upload-multiple-images-in-php-and-store-in-mysql/