如何使用两个浏览按钮和一个提交按钮上传两张不同图像
如何使用两个浏览按钮和一个提交按钮上传两张不同图像
提问于 2018-06-01 02:13:46
我想上传两张图片，一张是用户的，另一张是他的ID，使用mysqli的提交按钮。这是我的HTML。
<!DOCTYPE html>
<html>
<body>
<form action="/action_page.php">
your image: <input type="file" name="img"><br/>
your Id card: <input type="file" name="img2">
<input type="submit" name="publish" value="upload">
</form>
</body>
</html>所有我知道的是上传单个图像一次，但如果要上传这些图像到数据库中的单次提交怎么办。我不是在写PHP，因为我不知道怎么做。我可以使用一个数组一次上传多个图像，但我想使用这种方法。用PHP可以吗？？
PHP支持单次上传：
<?php
$dir = "uploads/";
$t_file = $dir . basename($_FILES["img"]["name"]);
$uploadOk = 1;
$imageFileType = strtolower(pathinfo($t_file,PATHINFO_EXTENSION));
if(isset($_POST["upload"])) {
$check = getimagesize($_FILES["img"]["tmp_name"]);
if($check !== false) {
echo "File is an image - " . $check["mime"] . ".";
$uploadOk = 1;
} else {
echo "File is not an image.";
$uploadOk = 0;
}
}
?>回答 1
Stack Overflow用户
发布于 2018-06-01 05:14:13
我可以用javascript来实现。所有尊敬的用户，请对那些不熟悉任何语言的人更有帮助。或者让这个网站只面向专家，而不是初学者。
因此，我使用PHP进行第一次上传，使用JS进行第二次图像上传。
下面是js：
<script>
function startUpload(){
document.getElementById('uploadProcess').style.visibility = 'visible';
document.getElementById('uploadForm').style.visibility = 'hidden';
return true;
}
function stopUpload(success,uploadedFile){
var result = '';
if (success == 1){
result = '<span class="sucess-msg">The file was uploaded successfully!<\/span><br/><br/>';
//Uploaded file preview
var embed = document.getElementById("UploadedFile");
var clone = embed.cloneNode(true);
clone.setAttribute('src',uploadedFile);
embed.parentNode.replaceChild(clone,embed);
}else {
result = '<span class="error-msg">There was an error during file upload!<\/span><br/><br/>';
}
document.getElementById('uploadProcess').style.visibility = 'hidden';
document.getElementById('uploadForm').innerHTML = result + '<label>File:<input name="myfile" type="file" size="30" /><\/label><label><input type="submit" name="submitBtn" class="sbtn" value="Upload" /><\/label>';
document.getElementById('uploadForm').style.visibility = 'visible';
return true;
}
</script>html：
<form action="upload.php" method="post" enctype="multipart/form-data" onsubmit="startUpload();">
<p id="uploadForm">
Select image to upload:
<input type="file" name="fileToUpload" id="fileToUpload">
<input type="file" name="myfile" id="fileToUpload1">
<input type="submit" value="submitBtn" name="submit">
</p>
</form>和upload.php：
<?php
$success = 0;
$uploadedFile = '';
//File upload path
$uploadPath = 'uploads/';
$targetPath = $uploadPath . basename( $_FILES['myfile']['name']);
if(@move_uploaded_file($_FILES['myfile']['tmp_name'], $targetPath)){
$success = 1;
$uploadedFile = $targetPath;
}
sleep(1);
?>
<?php
$target_dir = "uploads/";
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
$uploadOk = 1;
$imageFileType = strtolower(pathinfo($target_file,PATHINFO_EXTENSION));
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.";
$uploadOk = 0;
}
}
if ($uploadOk == 0) {
echo "Sorry, your file was not uploaded.";
// if everything is ok, try to upload file
} else {
if (move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file)) {
echo "The file ". basename( $_FILES["fileToUpload"]["name"]). " has been uploaded.";
} else {
echo "Sorry, there was an error uploading your file.";
}
}
?>页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/50630428
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