在PHP中动态生成更新和删除表单。 [英] Dynamically generate update and delete form in PHP.

问题描述

你好！感谢目前为止所有的帮助。当我的老板说我应该尽快更新时，他的意思是上周。所以当我上班时，我得到了很好的欢乐。开始我的一天的好方法....


回到我的问题：


我在Mysql中有一个文本：获取代码它在PHP中是：

[PHP]

$ id = intval（mysql_real_escape_string（$ _GET [''id'']））;

$ query =" SELECT * FROM jos_issues WHERE jos_issues.contentid =''$ id''ORDER BY text desc" ;;

$ result = mysql_query（$ query）;


while（$ row = mysql_fetch_array（$ result））{

echo" $ row [text]< BR>" ;;

}

[/ PHP]


我正在寻找的是一种方式来支付
a）按一下链接进行更新

b）动态打印出一张表格。


我所追求的是一种编辑和删除数据库中所有文本的方法。它们都与内容$ id相关联。所以我的表看起来像这样：id |文字| contentid


我尝试了这段代码，但我得到的是一张非常空白的页面，上面没有任何内容：


[PHP]<？包括（QUOT; C：/db.inc"）; error_reporting（E_ALL）;

ini_set（''display_errors''，''1''）;

// $ id = intval（mysql_real_escape_string（$ _GET ['' id'']））;

$ id = 12; echo" id = $ id";


$ query =" SELECT * FROM jos_issues WHERE jos_issues.contentid =''$ id''" ;;

$ result = mysql_query（$ query）;

$ num = mysql_numrows（$ result）;

// mysql_close（）;


$ i = 0;

while（$ i< $ num）{

$ text = mysql_result（$ result，$ i，" text"） ;

< form action =" updated.php" method =" post">

< input type =" hidden"命名= QUOT; ud_id"值= QUOT;< ;? echo $ id; ？>">

文字：< input type =" text"命名= QUOT; ud_text"值= QUOT;< ;? echo $ text; ？>">< br>

< input type ="提交" value =" Update">

< / form>


++ $ i;

}

？> [/ PHP]


如果你能帮我解决这个问题，请告诉我。今天我没有午餐。我必须在一点钟完成，或者我被搞砸了......



提前致谢。


Phopman

解决方案

id = intval（mysql_real_escape_string（

_GET [''id'']）） ;

query =" SELECT * FROM jos_issues WHERE jos_issues.contentid =''

Hi there! Thanks for all the help so far. When my boss said I should be up to date as soon as possible, he meant last week. So I got a good rollicking when I came to work. Nice way to start my day....

Back to my problem:

I have a Text in Mysql: The code for fetching it in php is:
[PHP]
$id = intval( mysql_real_escape_string( $_GET[''id''] ));
$query = "SELECT * FROM jos_issues WHERE jos_issues.contentid= ''$id'' ORDER BY text desc";
$result = mysql_query($query);

while($row= mysql_fetch_array($result)) {
echo"$row[text]<BR>";
}
[/PHP]

What I am looking for is a way to
a) press a link for update
b) dynamically print out a form for doing that.

All I am after is a way to edit and delete all the texts that are in my database. They are all linked to a content $id. So my table looks like this: id | text | contentid

I tried this code, but all I get is a very blank page with nothing on:

[PHP]<? include("c:/db.inc"); error_reporting(E_ALL);
ini_set(''display_errors'', ''1'');
//$id = intval( mysql_real_escape_string( $_GET[''id''] ));
$id = 12; echo "id = $id";

$query = "SELECT * FROM jos_issues WHERE jos_issues.contentid= ''$id''";
$result = mysql_query($query);
$num = mysql_numrows($result);
//mysql_close();

$i = 0;
while($i<$num) {
$text = mysql_result($result,$i,"text");
<form action="updated.php" method="post">
<input type="hidden" name="ud_id" value="<? echo $id; ?>">
Text: <input type="text" name="ud_text" value="<? echo $text; ?>"><br>
<input type="Submit" value="Update">
</form>

++$i;
}
?>[/PHP]

Let me know if you can help me with this one. There''s no lunch for me today. I have to have finished at one o'' clock, or I am screwed...


Thanks in advance.


Phopman

解决方案

id = intval( mysql_real_escape_string(

_GET[''id''] ));

query = "SELECT * FROM jos_issues WHERE jos_issues.contentid= ''

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