PHP: Undefined offset error

Question

The below-provided code is used to fill an array "data".

$query1="SELECT * FROM tab1, tab2 WHERE tab1.column1=tab2.column2;";
    $result1=DatabaseConnector::ExecuteQueryArray($query1);
    $data = array();
    $i = 0;
    foreach ($result1 as $row):
        $data[] = array(
            array($i,array("xxx",' EE112',$row['column3'],'FT445'),"2004-03-01 10:00","2004-03-01 14:00"));
        $i++;
    endforeach;

When I try reading data from the array, the error "Undefined offset: 1" occurs. The funny thing is that when I filled "data" array using $data = and not $data[] =, there was no error, just the last row was filled. The error is produced by line $bar = new GanttBar(..). I tried to substitute $row['column3'] by some string "xxx", but there was the same error.

for($i=0; $i<count($data); ++$i) {
        $bar = new GanttBar($data[$i][0],$data[$i][1],$data[$i][2],$data[$i][3]);
        $graph->Add($bar);
    }

I added the line var_dump($data); I'm using Zend Studio. Where can I see the output of var_dump? Sorry I'm newbie. I tried Debug Mode and also looked at Console. There is no message there. — Klausos Klausos
– Klausos Klausos, Commented May 17, 2012 at 9:30
Run the code and you're supposed to see what $data looks like. Then compare that with the way you iterate over its elements. — Niko
– Niko, Commented May 17, 2012 at 10:58

xdazz · Accepted Answer · 2012-05-17 09:30:39Z

3

Isn't

$data[] = array("xxx",' EE112',$row['column3'],'FT445',"2004-03-01 10:00","2004-03-01 14:00");

just enough, data structure should be as simple as possible.

Ok, if you want that structure.

$data[] = array(array("xxx",' EE112',$row['column3'],'FT445'),"2004-03-01 10:00","2004-03-01 14:00");

Then

foreach($data as $i => $var) {
    $bar = new GanttBar($i, $var[0], $var[1], $var[2]);
    $graph->Add($bar);
}

edited May 17, 2012 at 9:30

answered May 17, 2012 at 9:22

xdazz

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5 Comments

Klausos Klausos Over a year ago

No, I'm working with jpGraph Gantt Chart and it requires specific input format. Well, if I manually fill the array "data", then everything works fine.

Niko Over a year ago

The second bit of code does not work with the third one - $var will be again an array containing an array. Use the first way to fill $data with the loop at the end.

xdazz Over a year ago

@Niko, He wants $var[0] to be an array.

Klausos Klausos Over a year ago

Ups, both your solution and mine are fine if I do $data[] = array($i,array(... The error was caused by $data[] = array(array($i,array(...

xdazz Over a year ago

@KlausosKlausos Why do you need include $i in the $data, you could just use the $i in the second loop.

sprain · Accepted Answer · 2012-05-17 09:36:06Z

1

I think there is a very basic problem:

The foreach loop is creating a new subarray in $data for every row. But later on your code is trying to get three rows from each entry in $data - which never has been created. Therefore $data[$i][1] is always empty.

You might need to adjust your sql query to receive the data in the correct format from the beginning.

answered May 17, 2012 at 9:36

sprain

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2 Comments

Klausos Klausos Over a year ago

The problem is that the following code also is not fine: $bar = new GanttBar($i,"xxx","2010-10-10 10:05","2010-10-10 11:00"); Maybe there is the problem with count($data)? I don't understand why this code works for manually created "data" array: $data = array(array(...),array(...),array(...));

sprain Over a year ago

Show us an example of your manually created data and what you do with it. Then also show us what print_r($row) contains.

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