How to send the same variable from php to two different php pages using javascript

Question

I have these 4 pages and works fine, but now i need to send the "word" typed in the input field at (index.php) to another page (pag2.php) at the same time it sends to (pag1.php) using this javascript code at (script.php).

index.php

<form method="post">
    <input type="text" name="word" id="word" onkeyup="getSugest(this.value);">
</form>
<div class='search'><img ..... searching.gif></div>
<div id="sugest">
<div id="resultIndex"></div>
<div id="buttonsIndex">
<ul>
<?
for($i=1; $i<=page; $i++;){
echo "<li id='".$i."'>".$i."</li>";
}
?>
<ul>
</div>
</div>

script.js

    function getSugest(value){
    if(value != ""){
        if (callPage1(value) || callPage2(value)) {
         var data1 = callPage1(value);
         var data2 = callPage2(value);
         //var combinedData = combine as data1 and data2 as you want
         $("#sugest").html(combinedData); 
        } else {
         $("#sugest").html("Theres nothing in DB!"); 
        }
    }
}

function callPage1(value) {
    $.post("pag1.php",{word:value},function(data){
            if(data != ""){
                return data;
            }
            else{
                false;
            }
        });
}

function callPage2(value) {
    $.post("pag2.php",{word:value},function(data){
            if(data != ""){
                return data;
            }
            else{
                false;
            }
        });
}
$(document).ready(function(){
    function showLoader(){
        $('.search').fadeIn(200);
    }
    function hideLoader(){
        $('.search').fadeOut(200);
    };
    $("#buttonIndex li").click(function(){
        showLoader();

        $("#buttonIndex li").css({'background-color' : ''});
        $(this).css({'background-color' : '#D8543A'});

        $("#resultIndex").load("pag1.php?page=" + this.id, hideLoader);

        return false;
    });
    $("#buttonCar li").click(function(){
        showLoader();

        $("#buttonCar li").css({'background-color' : ''});
        $(this).css({'background-color' : '#D8543A'});

        $("#resultCar").load("pag2.php?page=" + this.id, hideLoader);

        return false;
    });
    $("#1").css({'background-color' : '#D8543A'});
    showLoader();
    $("#resultIndex").load("pag1.php?page=1", hideLoader);
    $("#resultCar").load("pag2.php?page=1", hideLoader);

});

pag1.php

$word = mysql_real_escape_string(addslashes($_POST['word']));
echo "Word is: ".$word;
//here is the php and mysql querys to return the result

pag2.php

$word = mysql_real_escape_string(addslashes($_POST['word']));
echo "Word is: ".$word;
//here is the php and mysql querys to return the result

I appreciate any help.

What do you mean "to another page at the same time"? What are you having problems with? Just add another $.post. As for: "is this the best way" I'd suggest not using inline JavaScript, and using $('#word').keyup to bind the event. — gen_Eric
– gen_Eric, Commented Apr 27, 2012 at 22:00
Thanks for correct me. I think now is best explained. I already tried "just add another $.post", but didnt work. Could you send me an example please? — André Oliveira
– André Oliveira, Commented Apr 27, 2012 at 22:10
I created the same $.post just like this one, making it two block of codes just the same, only changing the page name from pag1.php to pag2.php. And when i go to the pag2.php does not appear the word. I mean, the variable doesnt go to the pag2.php. Does not display erros. — André Oliveira
– André Oliveira, Commented Apr 27, 2012 at 22:29

gggrishna1 · Accepted Answer · 2012-04-27 22:06:13Z

2

Don't use inline functions, use jQuery's native methods instead:

$(function(){
   $('#word').keyup(function(){
      if(this.value) {
         $.post("pag1.php", { word: value }, function(data) {
            if(data){
               $("#suggest").html(data);
            } else {
               $("#suggest").html("Theres nothing in DB!");
            }
         });
      }
   });
});

answered Apr 27, 2012 at 22:06

gggrishna1

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2 Comments

André Oliveira Over a year ago

Thanks for answer my question. What do i need to put instead of onkeyup="pegarSugest(this.value);"

Armatus Over a year ago

You just remove that. $('#word').keyup(function(){... does the equivalent of onkeyup="... but in a cleaner manner.

Hüseyin BABAL · Accepted Answer · 2012-04-28 12:04:38Z

2

There is an example scenario here

You can implement a method for each call, and after calling all of them, you can combine their results

You can also have a look at my sample code for reproduce your scenario (modify it according to your needs)

function getSugest(value){
    if(value != ""){
        if (callPage1(value) || callPage2(value)) {
         var data1 = callPage1(value);
         var data2 = callPage2(value);
         //var combinedData = combine as data1 and data2 as you want
         $("#sugest").html(combinedData); 
        } else {
         $("#sugest").html("Theres nothing in DB!"); 
        }
    }
}

function callPage1(value) {
    $.post("pag1.php",{word:value},function(data){
            if(data != ""){
                return data;
            }
            else{
                false;
            }
        });
}

function callPage2(value) {
    $.post("pag2.php",{word:value},function(data){
            if(data != ""){
                return data;
            }
            else{
                false;
            }
        });
}

edited Apr 28, 2012 at 12:04

answered Apr 27, 2012 at 21:59

Hüseyin BABAL

15.6k5 gold badges53 silver badges74 bronze badges

17 Comments

André Oliveira Over a year ago

thanks for the answer.. the example you sent is very complex and will change much of my code, i would like to maintain the structure. Could you write an example of separating it in methods? cheers

André Oliveira Over a year ago

cucuzoa, i really appreciate your help, and i think its nearly there. When i changed my code for this one, and typed a word inside of the input box, it shows me only the message "Theres nothing in DB!". Do i need to change anything in the others pages? In the pag1.php and pag2.php i keep calling the variable "word" as i did before? If so we need some adjustment in the code. cheers man

Hüseyin BABAL Over a year ago

I updated the conditional logic in the code. && to || . So, this means, if there is valid response data at least one page, you will write it to page, else, you will show error message. Could you please try new version of code?

André Oliveira Over a year ago

Still the same. Only show the message: "Theres nothing in DB!"

Hüseyin BABAL Over a year ago

You got no data from both page?

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