Question

我只是尝试上传图片并将其位置插入数据库（SQLYog）。没有数据库连接，代码工作正常。但是当我尝试将它与SQL Yog链接时，操作页面根本不显示任何内容，并且没有任何内容插入到数据库表中。如果你能帮忙的话会很棒。这是我的代码。

form.php的

<html>
<head>
<script type="text/javascript">
function validate(){
var filevalue=document.getElementById("file").value;
var description=document.getElementById("description").value;
if(filevalue=="" || filevalue.length<1){
alert("Select File.");
document.getElementById("file").focus();
return false;
}
if(description=="" || description.length<1){
alert("File Description must not be blank.");
document.getElementById("description").focus();
return false;
}

return true;
}
</script>
</head>
<body >
<h2 align="center" >File Upload</h2>
<form action="file_upload.php" method="post"
enctype="multipart/form-data" onSubmit="return validate()" >
<table align="center" >
<tr>
<td><label for="file">File:</label></td>
<td><input type="file" name="file" id="file" /></td>
</tr>
<tr>
<td><label >File Description:</label></td>
<td><input type="text" name="description" id="description" /></td>
</tr>
<tr>
<td></td>
<td><input type="submit" name="submit" value="Submit" /></td>
</tr>
<table>
</form>
</body>
</html>

file_upload.php

<?php
include("connect.php"); //database connection
if ((($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/jpeg")
|| ($_FILES["file"]["type"] == "image/pjpeg"))
&& ($_FILES["file"]["size"] < 1000000))
{
if ($_FILES["file"]["error"] > 0)
{
echo "File Error : " . $_FILES["file"]["error"] . "<br />";
}
else {

echo "Upload File Name: " . $_FILES["file"]["name"] . "<br />";
echo "File Type: " . $_FILES["file"]["type"] . "<br />";
echo "File Size: " . ($_FILES["file"]["size"] / 1024) . " Kb<br />";
echo "File Description:: ".$_POST['description']."<br />";

if (file_exists("images/".$_FILES["file"]["name"]))
{
echo "<b>".$_FILES["file"]["name"] . " already exists. </b>";
}else
{
move_uploaded_file($_FILES["file"]["tmp_name"],"images/". $_FILES["file"]["name"]);

$loc="images/".$_FILES["file"]["name"];
$qu="insert into images.img(loc) values('$loc')";
mysql_query($qu,$con);
?>
Uploaded File:<br>
<img src="images/<?php echo $_FILES["file"]["name"]; ?>" alt="Image path Invalid" >
<?php
}
}
}else
{
echo "Invalid file detail ::<br> file type ::".$_FILES["file"]["type"]." , file size::: ".$_FILES["file"]["size"];
}
?>

数据库名称：image
表名：img
表字段：imgid（int primary key，auto incr），loc（varchar）

Answer 1

您必须设置数据库连接变量并按以下方式检查您的查询：

$qu="insert into image.img (imgid,loc) values (NULL,'$loc')";

PHP图片上传不起作用

1 个答案: