for loop question

rolwong

It really drives me crazy. I have the following script:
echo "<SCRIPT LANGUAGE='JavaScript'>\n";
echo "function update_student()\n";
echo "{\n";
echo "var selected_class_no = document.myForm.individual_class_no.value;\n";
for ($i=0;$i<$total_student;$i++) {
$db_class_no = $arr[$i]['class_no'];
echo "if (selected_class_no == '$db_class_no')\n";
echo "{\n";
$db_student_name[] = $arr[$i]['student_name'];
echo "alert ('$i');\n";
echo "}\n";

}
foreach($db_student_name as $tempdata) {
echo "alert ('$tempdata');\n";
}
$student_count = count ($db_student_name);
echo "alert ('$student_count');\n";
echo "}\n";
echo "</SCRIPT>\n\n";

When I echo "alert ('$i')", it only shows me 2 numbers, that means the if-statement only execute twice, but when i echo $student_count, it shows me 6 numbers in the array...>_<

majiclab

First of all, please explain what you are trying to do, as without that, we have no idea what is wrong with your code, as we do not know what it is supposed to do! 🙂

Secondly, if I didn't know better, it appears that you are trying to mix PHP and JavaScript. PHP is server-side and JavaScript is client-side, you cannot mix the two. You can use PHP to automatically generate JavaScripts, however, from what I see, it looks like you are trying to use a JavaScript if() statement to control some PHP code. All of your PHP code is executed before it anything reaches the browser, at which point JavaScript will kick in.

If you tell me what you are trying to do, I might be able to help you out, otherwise your code is just something that I can't really grasp since I don't know what it's doing.

Get back to me and I'll see what I can do!

-Percy

rolwong

Thanks, let me explain in more details. I am trying to make a drop down menu and a table. When the user select a class name from drop down menu, the table would show the student names of the corresponding class. That's why i need to mix javascript and php together, i don't want to let the user make a selection, then send data to server and reload the page to create the corresponding student name list.

rolwong

Now I have a drop down menu showing the classname. The above php+javascript code mean to create the corresponding student name list of the corresponding selected class.

class no student_name
1 Kelvin
1 Martin
2 Jean
2 Louis
3 Mary
3 Sisi

var selected_class_no is the class no that just selected by the user. Then with this number, I compare it with DB (all data has all been downloaded when the page is loaded).

for ($i=0;$i<$total_student;$i++) {
$db_class_no = $arr[$i]['class_no'];
echo "if (selected_class_no == '$db_class_no')\n";
echo "{\n";
$db_student_name[] = $arr[$i]['student_name'];
echo "alert ('$i');\n";
echo "}\n";

}

This script is to compare the user-selected class_no with DB data, if the user select "1", then Kevlin, Martin would push into a new array and then display. But when i echo $i, it does show if-statment only execute twice, but when i echo count($db_student_name), this array has contained all six student name)

majiclab

The problem with what you want to do is that PHP doesn't execute with JavaScript. The PHP is server-side, so when calling update_student() from JavaScript doesn't run any of that PHP code you put in there, because all of that PHP code has already been run.

You are in essence trying to do something like this:

<SCRIPT>
class_no = 2;
<?php
for ($i = 1; $i <= 3; $i++) {
echo "if (class_no == $i) {";
echo "    alert('$i');";
$class_numbers[] = $i;
echo "}";
}

foreach ($class_numbers as $class_number) {
echo "alert($class_number);";
}
?>
</SCRIPT>

The problem with that code is that this is what the browser will get!

<SCRIPT>
class_no = 2;
if (class_no == 1) {
    alert('1');
}
if (class_no == 2) {
    alert('2');
}
if (class_no == 3) {
    alert('3');
}
alert('1');
alert('2');
alert('3');
</SCRIPT>

That is because all PHP code is executed before anything goes to the user's browser. PHP is server-side, not client-side. You cannot mix the two together!! A bad analogy is that you are trying to listen to a CD on your computer, but the CD is sitting in your car. I know that analogy doesn't really make sense... but the point is: You can't listen to a CD on your computer when the CD is sitting in your car. You have to have the CD in your computer to listen to it in your computer.

You can't use an if() statement in JavaScript to conditionally executed some PHP code, because the PHP code is executed before the JavaScript code is ever seen by the browser.

If you right click on your page, and click on "View Source", you will see what I am talking about. The alerts are hardcoded in the HTML page. When you view source on your computer, you don't see any PHP code in it, that's be cause it is server-side.

I hope you are understanding what I am saying... There is a way to do what you are wanting to do, but you're trying to do it in a way that is not possible.

Hope that helps,

-Percy

rolwong

Thanks for your comment first. I'd like to hear some suggestions on what I do:

Let say, i have a school with 20 classes, each class has 30 students. I want to do a voting system for the whole school. User can select any students from any class for their good performance. What I did before was create a html table to list out all student at one time. But you know, the list will be very very long and not user-friendly. That is why I was thinking to allow user to select class first, then showing student names of corresponding class and then vote. It involved javascript and PHP. It makes me stuck.

I am now thinking to divided the long student list to several pages. But what I am guessing is: when the user has voted at the first page, then want to see the student list at second page, it will reload the page to show the second page student list, this is fine. But the browser can no longer remember what he has voted from the first page. So I am thinking is there any possible method to remember the variable even though the page is reloaded.

majiclab

Couldn't you just make them vote on the second page, when they select a student?

-Percy

rolwong

It may be the only solution till now.