Question

我需要一些建议/帮助。我正在尝试上传图片，但是wampserver不断向我显示此错误（注意：未定义的变量：第32行的C：\ wamp \ www \ Shopaholic \ upload_file.php中的名称），如果有人可以在这里提供帮助，我将不胜感激。感谢

这是我的代码

<form action="upload_file.php" method="post" enctype="multipart/form-data">
<label for="file">Filename:</label>
<input type="file" name="file" id="file"><br>
<input type="submit" name="submit" value="Submit">
</form>




<?php



if(isset($_POST['submit'])){
$name       = $_FILES['file']['name'];  
$temp_name  = $_FILES['file']['tmp_name'];  
if(isset($name)){
    if(!empty($name)){      
        $location = '..Shopaholic/upload/';      
        if(move_uploaded_file($temp_name, $location.$name)){
            echo 'uploaded';
        }
    }       
}  else {
    echo 'please uploaded';
}
}


$sql1= mysql_query("INSERT INTO dumimage (name)values('$name')");

?>

这是错误行32：$sql1= mysql_query("INSERT INTO dumimage (name)values('$name')");

Answer 1

您需要将查询包装在IF语句中。如果它不存在，您就无法使用它。在您的文件上传并确认为成功之后，我会把它放进去。（也可能要检查你的IF语句。不确定所有这些都是必要的。）

<?php
$conn = mysql_connect("localhost","root","") or die (mysql_error());
mysql_select_db("shopaholic",$conn) or die (mysql_error());

if(isset($_POST['submit'])){

    $name       = $_FILES['file']['name'];  
    $temp_name  = $_FILES['file']['tmp_name'];  
    if(isset($name)){
        if(!empty($name)){      
            $location = '..Shopaholic/upload/';      
            if(move_uploaded_file($temp_name, $location.$name)){
                echo 'uploaded';
                $sql1= mysql_query("INSERT INTO dumimage (name)values('$name')"); 
            }
        }       
    }  else {
        echo 'please uploaded';
    }
}

?>

将图像上传到wampserver时出现未定义的变量错误

1 个答案: