不知道如何解决（在PHP和Javascript之间） [英] Don't know how to solve (Between Php and Javascript)
问题描述
我不知道如何通过以下脚本解决变量 $ cat 。 text 从Form到javascript的变量，并传递给php函数为 $ Categories_name ，并成为 $ cat 。
我已经测试了 $ cat 变量，它不是String，它是对象，我不明白。
但是我需要 $ cat 为String。
When Test =这是猫（3字），
我用php str_word_count测试$ cat，输出为1（我需要更正答案3）;
我测试$ cat by php var_dump并且没有输出（我需要更正答案String）。
< p id =CaTable>< / p>
< script>
函数CaFunction（）
{
var text = document.getElementById（CategorySelect）。value;
document.getElementById（CaTable）。innerHTML =<？php php_catable（'+ text +'）;？>;
}
< / script>
<！ - 通过php和MySQL生成表格 - >
<？php
函数php_catable（$ Categories_name）
{
$ cat = $ Categories_name;
.................
.................
$ sql = select * from table where xyz ='。$ cat。 ;
}
？>
您的第一页：
假设 CategorySelect 是一个下拉选择框，为其onChange事件创建一个脚本，并创建一个 method =post隐藏的输入到generate_table.php。
< input type =hiddenname =ca_tableid = ca_table/>
使ca_table成为一个隐藏的输入，以便在提交此页时php会从中获取值
< script language =javascripttype = text /的javascript>
函数CaFunction（）{
documentGetElementById（'ca_table'）。value = documentGetElementById（'CategorySelect'）。value;
submit（）;
}
< / script>
将其添加到您的选择下拉菜单中：
onChange =CaFunction（）;
您的接收页面：
所以您的接收页面generate_table.php p>
<？php
函数php_catable（$ Categories_name）
{
$ cat = $ Categories_name ;
.................
.................
$ sql = select * from table where xyz ='。$ cat。 ;
}
$ category_name = $ _POST ['ca_table']; //至少用合适的preg_replace等清理
//并调用你的可捕获函数
php_catable（$ category_name）;
？>
这样，您的结果将按照客户端/服务器的注释发回服务器通过 @Fluinc 并通过 @litelite 。为了让它做一些看起来像innerHTML的东西，它改变了页面的一部分而不提交整个页面，你将需要AJAX，同样是按照 @litelite 的答案。
可能因为依赖JavaScript而被标记，但主要是为了帮助澄清客户端v服务器。
如果您想要避免此脚本的JavaScript依赖性，您可以完全省略onChange并添加一个提交按钮，然后收集 $ _ POST ['CategorySelect']; 假设它是它的名字 - 确保它有
为了使AJAX具有可视化效果（尽管页面仍然提交），您可以将页面提交给自己在表单上使用 action =<？php echo $ _SERVER ['PHP_SELF'];？>，并将所有代码放在一页上。您可以将表格生成代码放在希望显示表格的div中 - 当然，它需要一个默认状态集合。
@litelite 对不直接使用发布数据的评论在一个sql查询中，对于防止攻击也很重要 - 确保在使用它之前清理它！
I don't know how to solve the variable $cat by following script.
"text" variable from Form to javascript and pass to php function to be $Categories_name and to be $cat.
I already test the $cat variable, it is not "String", it is "object", I don't understand.
But I need $cat to be "String".
when Test = "This is Cat" (3 words),
I test $cat by php str_word_count and the output is 1 (I need to correct answer 3);
I test $cat by php var_dump and no output (I need to correct answer "String").
<p id="CaTable"></p>
<script>
function CaFunction()
{
var text = document.getElementById("CategorySelect").value;
document.getElementById("CaTable").innerHTML = "<?php php_catable('" + text + "'); ?>";
}
</script>
<!-- Generate Table by php and MySQL-->
<?php
function php_catable($Categories_name)
{
$cat = $Categories_name;
.................
.................
$sql = "select * from table where xyz = '" .$cat. "'";
}
?>
Your First Page:
Assuming CategorySelect is a dropdown select box, create a script for its onChange event and create a method="post"post form with a hidden input that goes to "generate_table.php".
<input type="hidden" name="ca_table" id="ca_table" />
You make ca_table a hidden input so php will pick up the value from it when this page gets submitted to a second page where you can generate your table using the php function.
<script language="javascript" type=text/javascript>
function CaFunction(){
documentGetElementById('ca_table').value = documentGetElementById('CategorySelect').value;
submit();
}
</script>
add this to your select dropdown:
onChange="CaFunction();"
Your Receiving Page: So your receiving page "generate_table.php" would have
<?php
function php_catable($Categories_name)
{
$cat = $Categories_name;
.................
.................
$sql = "select * from table where xyz = '" .$cat. "'";
}
$category_name = $_POST['ca_table']; // cleaned up at least with suitable preg_replace etc
// and call your catable function
php_catable($category_name);
?>
So that way your result will have been posted back to the server as per comments about client side/server side by @Fluinc and answer by @litelite. To get it to do something which performs looking like innerHTML which changes a part of the page without submitting the whole page you will need AJAX, again as per @litelite's answer.
Might get marked down for being dependant on JavaScript but intended mostly to help clarify client v server.
If you want to avoid the JavaScript dependency of this script you could leave out the onChange altogether and add a submit button, then collect $_POST['CategorySelect']; assuming that is its name - ensure it has name="CategorySelect" for php as well as its Id for your css/javascript. Php gets its variable from the item's name.
To get something a bit like the effect of AJAX visually (though the page is still submitted) you could submit the page to itself using action="<?php echo $_SERVER['PHP_SELF']; ?>" on the form and have all the code on the one page. You can put the table generating code in the div where you want the table to appear - it would need a default state set, of course.
@litelite's comment regarding not using posted data directly in an sql query is also vital to prevent attack - make sure you clean it up before you use it!
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