我在这里使用此 php 代码在数据库表的一个字段中插入多个图像，但在插入一行后每次插入后都会多一行.......帮我解决这个问题.. 谢谢提前。
<?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>
最佳答案
请检查表单是否已发布然后执行插入，否则在加载页面时将在表中插入一行数据为空
<?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."')");
}
?>
关于php - 多个图像插入在每次插入后额外占用一行，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/27540238/