Question

该法典行之有效。我可以说明如何将数据插入db 如果用户首次使用“SAVE”纽吨或更新数据。

www.un.org/Depts/DGACM/index_spanish.htm php一侧

<?php
require  ../../core/includes/common.php ;

$name=filter($_POST[ name ], $db);
$title=filter($_POST[ title ], $db);
$parentcheck=filter($_POST[ parentcheck ],$db);
if(isset ($_POST[ parent ])) $parent=filter($_POST[ parent ],$db);
else $parent=$parentcheck;  
$menu=filter($_POST[ menu ], $db);
$content = $db->escape_string($_POST[ content ]);

$result=$db->query("INSERT INTO menu (parent, name, showinmenu) VALUES ( $parent ,  $name ,  $menu )") or die($db->error);
$new_id = $db->insert_id;
$result2=$db->query("INSERT INTO pages (id, title, content) VALUES ( $new_id ,  $title ,  $content )") or die($db->error);  

if ($new_id>0){  
echo "{";  
echo  "msg": "success"  ;
echo "}";  
}else{ 
echo "{"; 
echo
 "err": "error" ;  
echo "}";  
}

?>

UPDATE

由于@jmlsteeke i 找到了途径

将这部法典放在html部分

<?php
$result=$db->query("INSERT INTO menu (parent, name, showinmenu) VALUES ( 555 ,  new ,  0 )") or die($db->error);
$new_id = $db->insert_id;
$result2=$db->query("INSERT INTO pages (id, title, content) VALUES ( $new_id ,  new ,  new )") or die($db->error);  

?>

添加以下文字:

<input type="hidden" name="id" value="<?=$new_id?>"/>

使用的服务器代码

$result=$db->query("UPDATE pages AS p, menu AS m SET m.parent= $parent , m.name= $name , m.showinmenu= $menu , p.id= $id , p.title= $title , p.content= $content  WHERE m.id= $id  AND p.id=m.id") or die($db->error);

感谢@jmlsteeke

Answer 1

一种常见的做法是,在你编辑该页时,把id作为隐蔽的场所。当用户提交该页时,如果在场的话,你就会发布统一民主党的指挥,如果在场的话,你会知道这个新页,并发布INSERT的指挥。

如果你需要我更透彻的话,让我知道。

Edit: 更富裕

I ll make a simple, complete, example of what I mean.

Form.php pseudo Code

//set default values for fields
//print form tag
if (isset($ id ,$_GET)) {
    //fetch data from database
    //print hidden id field
    //override default values for fields
}
//print rest of fields using default values (possibly overridden)

DoForm.php pseudo Code

//Sanitize user input
if (isset( id ,$_POST)) {
    //UPDATE database with user input
} else {
    //INSERT new rows into table with user input
}

请允许我说,有一个叫作表.php的网址,负责显示表格,另一个称为DoForm.php的网址,负责处理表格。

如果用户访问表.php没有指定身份(http://example.com/Form.php),就会显示以下表格:

<form method="post" action="DoForm.php">
<input type="text" name="name" value="" />
<input type="text" name="title" value="" />
... other stuff ...
</form>

用户将添加一些信息,点击提交纽特和多Form,将获得以下持久性有机污染物:

"name"  => "NewPageName"
"title" => "My First Webpag" [intetional typo, see later]
... other stuff ...

表格将检查是否存在<代码>_POST[id]。性别问题专家咨询组发布新页。

Later on, the user realises the made a typo, and goes to fix it. The user clicks on the "Edit Page" control for "NewPageName" which will be http://example.com/Form.php?id=1

表格. .php 参看id,其印刷形式如下:

<form method="post" action="DoForm.php">
<input type="hidden" name="id" value="1"
<input type="text" name="name" value="NewPageName" />
<input type="text" name="title" value="My First Webpag" />
... other stuff ...
</form>

用户固定其类型,将网络空间改为网页,并提交点击。 • 收到以下员额变量

"id"    => 1
"name"  => "NewPageName"
"title" => "My First Webpage"
... other stuff ...

DoForm sees that id is set, and so uses UPDATE instead of INSERT.

更加清楚吗?

Answer 2

MySQL has an INSERT ... ON DUPLICATE KEY UPDATE feature that will let you try to insert a row, or fall back to an update if it discovers a duplicate key (i.e. the row already exists).

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