...我想请问下,当我点击修改的按钮时,我用echo "dddddddddddddddd"打印字符串,为什么页面没有显示??;这样写修改的方法有问题吗????
if ($_POST[sumbit]) { echo "dddddddddddddddd";
$id = $_POST['id'];
$category_id = $_POST['category_id'];
$record = array ();
$record = $category_id;
$edit = $con->update($table, $record, $id);
if ($edit) {
echo "修改成功";
} else {
echo "修改失败";
}
}if (!empty ($id)) {
$sql = "select * from picture where id='$id'";
$rows = $con->query($sql);}
<form action="update.php" method="post"><table broder="1"><tr><th>编号:</th><th>
<input type="text" name="id" value="<?php echo $rows[0][id]?>" size="40" maxlength="40" /></th></tr>
<tr><th>类别编号:</th><th> <input type="text" name="category_id" value="<?php echo $rows[0][category_id]?>" size="40" maxlength="40"/></th></tr><input type="submit" name="submit" value="修改"/>
</from>
php.ini register_global 是不是off
php.ini register_global 是不是off
$_POST['submit'] 不是 $_POST[sumbit]
错误!~~
echo "数据已经提交。"
}else{
echo "你没有提交数据?"
}
我想问下,$pic = $_POST['pic']; 我打印echo $pic没有数据..我的数据库存的是图片名如 aa.jpg
<img src="../images/<?php echo $rows[0][pic]?>">
我想要取得图片名应该怎么取??
$img=$_POST['imgname'];
应该是$pic = $_POST[pic]这样... ..刚试了,有取到图片的名子
$table = 'picture';
if ($_POST[submit]) {
$id = $_POST['id'];
$category_id = $_POST['category_id'];
$name = $_POST['name'];
$pic = $_POST['pic'];
$description = $_POST['description'];
$record = array ($category_id,$name,$pic,$description);
print_r($record);
if ($con->update($table, $record, $id)) {
echo "修改成功";
return true;
} else {
echo "修改失败";
return false;
}
}print_r($record)得到数:Array ( [0] => 1 [1] => 凤惑天下fff [2] => 206013.jpg [3] => 简介省略 ) ...已经取到值了,为什么不可以修改..这是mysql类
public function update($tablename, $values, $id)
{
$comma = '';
foreach($values as $key=>$value){
$v = $comma . $key . "= '".$value."'";
$comma = ' AND ';
}
$sql = "UPDATE `$tablename` SET ".$v." WHERE id = $id";
return mysql_query($sql) ? true : false;
}
Array ( [0] => 1 [1] => 凤惑天下fff [2] => 206013.jpg [3] => 简介省略 )
这样的数据哪知道你的数组每一项对应数据库哪个字段啊?
"category_id" => $category_id,
"name" => $name,
"pic" => $pic,
"description"=>$description
);echo $fruits['name'];