Question

这里我使用这个php代码在数据库表的一个字段中插入多个图像，但插入后每行插入一行后再加一行.......帮我解决这个问题.. 谢谢你提前。

<?php
mysql_connect("localhost","root","");
mysql_select_db("test");
/*if(isset($_REQUEST['submit']))
{


         $pname=$_FILES['image']['name'];
         $tmp_name=$_FILES['image']['tmp_name'];
         move_uploaded_file($tmp_name."photo/".$pname);

        $fileext = pathinfo($pname, "photo/");
        $fileext = strtolower($fileext);

}*/

$uploads_dir = 'photo/';
//$images_name ="";
    foreach ($_FILES["image"]["error"] as $key => $error) {
        if ($error == UPLOAD_ERR_OK) {
            $tmp_name = $_FILES["image"]["tmp_name"][$key];
            $name = $_FILES["image"]["name"][$key];
            move_uploaded_file($tmp_name, "$uploads_dir/$name");
            $images_name =$images_name.$name.",";
        }
    }

    $sql=mysql_query("INSERT INTO multiimg(image) values('".$images_name."')");


?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
<script>
function addmore(num)
{
    if(num==1)
    {
        document.getElementById('field2').style.display='block';
        document.getElementById('ni1').style.display='block';
        return false;
    }
    else if(num==2)
    {
        document.getElementById('field3').style.display='block';
        return false;
    }

}
</script>
</head>

<body>
<form enctype="multipart/form-data" name="" action="" method="post">
    <div id="field1">Enter One Image :<input type="file" name="image[]" id="img1"/><a href="#" onclick="addmore(1)" id="ni1">addmore...</a></div>
    <div id="field2"  style="display:none;">Enter Two Image :<input type="file" name="image[]" id="img2"/><a href="#" onclick="addmore(2);">add more...</a></div>
    <div id="field3"  style="display:none;">Enter Three Image :<input type="file" name="image[]" id="img3"/><a href="#" onclick="addmore(3)" id="ni3">addmore...</a></div>
    <div id="field4" style="display:none">Enter Forth Image :<input type="file" name="image[]" id="img4"/><a href="#" onclick="addmore(4)" id="ni4">addmore...</a></div>

    <input type="submit" name="submit"/>

</form>

</body>
</html>

Answer 1

请检查表单是否已发布然后执行插入，否则在加载页面时，行将插入带有空数据的表中

<?php
    mysql_connect("localhost","root","");
    mysql_select_db("test");

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

    $uploads_dir = 'photo/';
    //$images_name ="";
        foreach ($_FILES["image"]["error"] as $key => $error) {
            if ($error == UPLOAD_ERR_OK) {
                $tmp_name = $_FILES["image"]["tmp_name"][$key];
                $name = $_FILES["image"]["name"][$key];
                move_uploaded_file($tmp_name, "$uploads_dir/$name");
                $images_name =$images_name.$name.",";
            }
        }

        $sql=mysql_query("INSERT INTO multiimg(image) values('".$images_name."')");

    }
    ?>

每次插入后多个图像插入需要额外的一行

1 个答案: