所以我已经在这个学校项目上工作了大约一个星期，然后遇到了一个实例，我必须实现一个用户注册表单，它应该提交用户信息以及用户配置文件图像的路径是也存储在DB中。缺点是我似乎无法使上传脚本正常工作。我已经观看了数十个教程，并从这个平台上阅读了类似问题的15个解决方案，但仍无济于事。我还确保php.ini文件打开了file_upload设置。
这是我的HTML表格
<form style="padding: 1em;" class="ui form small inverted raised segment" action="signup.php" method="POST" enctype="multipart/form-data">
<h4 class="ui dividing header">Background Information</h4>
<div class="field">
<div class="three fields">
<div class="field">
<input type="text" name="fname" placeholder="Firstname" required>
</div>
<div class="field">
<input type="text" name="mname" placeholder="Other name">
</div>
<div class="field">
<input type="text" name="lname" placeholder="Lastname" required>
</div>
</div>
</div>
<div class="field">
<div class="two fields">
<div class="field">
<select name="gender">
<option value="">Select Gender</option>
<option value="Male">Male</option>
<option value="Female">Female</option>
<option value="Other">Other</option>
</select>
</div>
<div class="field">
<input type="date" name="dob" required>
</div>
</div>
</div>
<div class="ui dividing header">Contact Information</div>
<div class="field">
<div class="two fields">
<div class="field">
<input type="tel" name="phone" placeholder="Mobile phone number" required>
</div>
<div class="field">
<input type="email" name="email" placeholder="Email address" required/>
</div>
</div>
</div>
<div class="field">
<div class="three fields">
<div class="field">
<input type="text" name="area" placeholder="Area/Village" required>
</div>
<div class="field">
<input type="text" name="trad_auth" placeholder="T/A or STA">
</div>
<div class="field">
<select id="district" name="kasungu">
<option value="Kasungu">Kasungu</option>
</select>
</div>
</div>
<div class="ui dividing header">Work Details</div>
<div class="field">
<div class="two field">
<div class="field">
<select id="department" name="department">
<option value="">Select Department</option>
</select>
</div>
</div>
</div>
<div class="ui dividing header">Acc Authentication Details</div>
<div class="field">
<input type="text" name="activation-code" placeholder="Enter admin authentication code (XXX-XXXX-XXXX)" required/>
</div>
<div class="field">
<div class="two fields">
<div class="field">
<input type="password" name="pass1" placeholder="Create Password" required/>
</div>
<div class="field">
<input type="password" name="pass2" placeholder="Confirm Password" required/>
</div>
</div>
</div>
<div class="ui dividing header">Upload Image</div>
<div class="field">
<div class="two fields">
<div class="field">
<input type="file" style="display: none;" id="pic" name="image"/>
<a type="button" id="upload" class="ui button fluid negative mini"><i class="icon camera"></i></a>
</div>
</div>
</div>
<div class="three fields">
<div class="field"></div>
<div class="field">
<button type="submit" name="submit" class="ui fluid mini positive button">
Signup <i class="icon user"></i>
</button>
</div>
<div class="field"></div>
</div>
<div>
<?php
if(!empty($errorMsg)){
echo $errorMsg;
}
?>
</div>
</div>
</form>
这是我接收FILE['image'] A PARAMETER的PHP功能。
<?php
function uploadFile($file){
$newName;
if(isset($file)){
$file_name = $file['name'];
$file_tmp_loc = $file['tmp_name'];
$file_size = $file['size'];
$file_error = $file['error'];
$ext = strtolower(end(explode('.', $file_name)));
$allowed = array('jpg','jpeg','png','gif');
if(in_array($allowed, $allowed)){
if($file_error == 0){
if($file_size > 3000000){
$file_name_new = uniqid('',true).".".$ext;
$destination = '../images/users/';
$destination = $destination.$file_name_new;
if(move_uploaded_file($file_tmp_loc, $destination)){
$newName = $destination;
}else{
echo "<h1>Failed to move uploaded file</h1>";
}
}
}else{
echo "file upload failed. ".$file['error'];
}
}else{
echo "Type not allowed!"; }
}
}
return $newName;
?>
在我的其他PHP脚本中，我调用的函数如下：
<?php
$fileName = uploadFile($_FILE['image']);
?>
然后，将返回的新名称存储在MySQL DB中作为其引用
你收到文件时必须使用$ _FILES，第一个数组是$ _FILES ['你在输入文件中提到的文件名，如name ='abc'']，第二个数组是关于文件名，大小，错误，位置。你也可以通过使用print_r（$ _ FILES ['image']）来看到这个;
<input type="file" style="display: none;" id="pic" name="image"/>
$file_name = $_FILES['image']['name'];
$file_tmp_loc = $_FILES['image']['tmp_name'];
$file_size = $_FILES['image']['size'];
$file_error = $_FILES['image']['error'];
所以纠正它！
为你加油！
这是运行代码。
$image_file = $_FILES['school_logo']['name'];
$tmp=$_FILES["school_logo"]["tmp_name"];
$ext = pathinfo($image_file ,PATHINFO_EXTENSION);
$folder="images/school_logo/";
$logos='logo'.'-'.uniqid(date('dmy')).rand().'.'.$ext;
$test=move_uploaded_file($tmp,$folder.$logos);
我希望这段代码对你有所帮助。
正如我所看到的，文件的输入字段名称是image。所以你应该将一个文件对象传递给被调用的函数$_FILES。
<?php
$fileName = uploadFile($_FILES);
?>
查看输入文件字段
<input type="file" style="display: none;" id="pic" name="image"/>
并更改您尝试上传图片的代码：
$file_name = $file['image']['name'];
希望它会对你有所帮助。