【发布时间】:2016-05-23 10:14:25
【问题描述】:
我正在做一个上传图片的页面,一切都很好。但是当我不小心上传图片而没有选择任何图片时,问题就出现了。出现警告“警告:file_get_contents():文件名不能为空”。我尝试解决,但仍然出现警告。我需要添加什么吗?我应该怎么做才能删除该警告?这是我的php代码
<form name="form1" action="" method="post" enctype="multipart/form-data">
<table>
<tr>
<td>select files</td>
<td><input type="file" name="f1"></td>
</tr>
<td><input type="submit" name="submit1" value="upload"></td>
<td><input type="submit" name="submit2" value="display"></td>
</table>
</form>
<?php
$con = mysqli_connect("localhost","root","");
mysqli_select_db($con,"image")or die(mysqli_error($con));;
if(isset($_POST["submit1"]))
{
$image = addslashes(file_get_contents($_FILES['f1']['tmp_name']));
$sql ="INSERT INTO images (username, image) VALUES('$_SESSION[username]','$image')";
$result=mysqli_query($con, $sql);
}if(isset($_POST["submit2"]))
{
$sql = "SELECT username, image FROM images where username = '$_SESSION[username]' ";
$result = mysqli_query($con,$sql);
echo"<table>";
echo "<tr>";
while ($row=mysqli_fetch_assoc($result))
{
echo "<td>";
echo '<img src="data:image/jpeg;base64,'.base64_encode($row['image']).'" height="100" width="100"/>';
echo "</td>";
}
echo "</tr>";
}
?>
?>
【问题讨论】:
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也发布您的表单代码!!
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阅读this以检查图像验证