ansaurus

Question

Answer 1

+5 A:

Just populate your global data structure directly rather than passing it through a JavaScript function. You are likely running into a variable scope problem since you are allocating memory in addEmployee.

Example:

<?php
print "details[details.length] = new Array('$id1', '$name1', '$salary1');\n";
?>

Will Bickford 2008-12-25 03:23:58

Answer 2

+1 A:

It looks like you're missing some curly braces in your for loop. Right now, you've got a 1-line for loop that does nothing but assign a value to your temp variable, which is re-declared on every iteration. I can't remember what the scope rules are for JavaScript in this situation, but it's generally bad practice to do this (if its intentional). Try something like this for your displayEmployee() method.

function displayEmployee(){
    for(var i=0;i<details.length;i++) {

        var temp_details= details[i];

        //display the id,name,sal        
        var id=temp_details[0];
        var name=temp_details[1];
        var sal=temp_details[2];

        alert(id);
        alert(name); 
        alert(sal);
    } 
}

That should fix your immediate problem. If you have control over all of your code, I'd recommend turning the employee data into an object so it's more manageable. To do this, you should use this PHP code:

<?php
echo "<SCRIPT LANGUAGE='javascript'>addEmployee({id:'".$id1."', name: '".$fname1."', salary:".$sal1."});</SCRIPT>";
// ...
echo "<SCRIPT LANGUAGE='javascript'>addEmployee({id:'".$id2."', name: '".$fname2."', salary:".$sal2."});</SCRIPT>";
// ...
?>

And then alter your JavaScript to look like this:

var details=new Array();

function addEmployee(employeeObj){
 details[details.length]=employeeObj;

 alert(employeeObj);
}
function displayEmployee(){
    for(var i=0;i<details.length;i++) {
        alert(details[i].id);
        alert(details[i].name); 
        alert(details[i].salary);
    } 
}

I haven't tested this code but it should run.

Dan Herbert 2008-12-25 03:33:42

Answer 3

A:

You have an error in your javascript call generated by PHP

echo "<SCRIPT LANGUAGE='javascript'>add dd_Employee('".$id1."','".$fname1."',".$sal1.");</SCRIPT>";

should be

echo "<SCRIPT LANGUAGE='javascript'>addEmployee('".$id1."','".$fname1."',".$sal1.");</SCRIPT>";

And if you want to change it to OOP?

<?php
$employeeList[] = array('sdf3434', 'Bob', 123.34);
$employeeList[] = array('wer335dg', 'Joe', 223);
?>
<html>
<head>
<script> 
function Employees() {
    this.list = new Array();
}
Employees.prototype.add = function (id, fname, sal) {
    this.list.push(new Array(id, fname, sal));
};
Employees.prototype.display = function () {
    for(var i=0, t = this.list.length;i<t;i++) {
        var emp = this.list[i];
        alert(emp[0]);
        alert(emp[1]); 
        alert(emp[2]);
    } 
};
var employees = new Employees();
<?php
foreach ($employeeList as $employee) {
    list($id, $fname, $sal) = $employee;
    printf("employees.add('%s', '%s', %f);\n", $id, $fname, $sal);
}
?>
</script>
</head>
<body>
<a href="#" onclick="employees.display();">click</a>
</body>
</html>

OIS 2008-12-25 22:49:58

Answer 4

A:

I have added some info regarding using a JSON object for this in another question. Hope it helps, http://stackoverflow.com/questions/393479/best-way-to-transfer-an-array-between-php-and-javascript

barfoon 2008-12-26 05:43:02

Answer 5

A:

This can be helpful for some: http://www.satya-weblog.com/2007/03/passing-value-javascript-php-javascript.html

2009-09-02 04:50:43

ansaurus

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Passing Value to Javascript from Php

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