用数组中的值填充PHP list（）

Question

I have an array:

$arr = array('foo', 'bar', 'bash', 'monkey', 'badger');

I want to have the elements in that array appear as the variables in my list():

list($foo, $bar, $bash, $monkey, $badger) = $data;

Without actually specifying the variables, I tried;

list(implode(",$", $arr)) = $data; and
list(extract($arr)) = $data;

But they don't work, I get:

Fatal error: Can't use function return value in write context

Does anyone have any idea whether this is possible?

---

UPDATE: more context:

I am getting a CSV of data from an API, the first row is column names, each subsequent row is data. I want to build an associative array that looks like this:

$data[0]['colname1'] = 'col1data';
$data[0]['colname2'] = 'col2data';
$data[0]['colname3'] = 'col3data';
$data[1]['colname1'] = 'col1data';
$data[1]['colname2'] = 'col2data';
$data[1]['colname3'] = 'col3data';

Before I do that, however, I want to make sure I have all the columns I need. So, I build an array with the column names I require, run some checks to ensure the CSV does contain all the columns I need. Once thats done, the code looks somewhat like this (which is executed on a foreach() for each row of data in the CSV):

//$data is an array of data WITHOUT string indexes
list(   $col1,
    $col2,
    $col3,
    ...
    $col14
 ) = $data;

foreach($colNames AS $name)
{
    $newData[$i][$name] = $$name;
}
// Increemnt
$i++;

As I already HAVE an array of column name, I though it would save some time to use THAT in the list function, instead of explicitly putting in each variable name.

The data is cleaned and sanitised elsewhere.

Cheers,

Mike

Answer 1

I want to have the elements in that array appear as the variables in my list():

i think there is your problem in understanding. list() does not create a new list structure or variable, it can only be used to assign many variables at once:

$arr = array(1, 2, 3);
list($first, $second, $third) = $arr;
// $first = 1, $second = 2, $third = 3

see http://php.net/list for more information.

you are probably looking for an associative array . you can create it with the following code:

$arr = array('first' => 1, 'second' => 2, 'third' => 3);
// $arr['first'] = 1, …

Answer 2

If some rows in your input file are missing columns, you can't really know which one is missing. Counting the number of values and aborting or jumping to next row when less than expected should be enough.

... unless you set the rule that last columns are optional. I'll elaborate on this.

Your code sample is far for complete but it seems that your problem is that you are using arrays everywhere except when matching column names to cell values. Use arrays as well: you don't need individual variables and they only make it harder. This is one of the possible approaches, not necessarily the best one but (I hope) clear enough:

<?php

$required_columns = array('name', 'age', 'height');
$input_data = array(
    array('John', 33, 169),
    array('Bill', 40, 180),
    array('Ashley', 21, 155),
    array('Vincent', 13), // Incomplete record
    array('Michael', 55, 182),
);
$output = array();

foreach($input_data as $row_index => $row){
    foreach($required_columns as $col_index => $column_name){
        if( isset($row[$col_index]) ){
            $output[$row_index][$column_name] = $row[$col_index];
        }
    }
}

print_r($output);

?>

I've used $row_index and $col_index for simplicity.

Original answer, for historical purposes only ;-)

I can't really understand your specs but PHP features variable variables :

<?php

$arr = array('foo', 'bar', 'bash', 'monkey', 'badger');

foreach($arr as $i){
    $$i = $i;
}

?>

Now your script has these variables available: $foo, $bar... It's quite useless and potentially dangerous but it does what you seem to need.

Answer 3

why not immediatly call the vars like

$arr['foo']

or

$arr[0]

Answer 4

You are trying to use a language construct in a manner in which it's not meant to be used. Use variable variables as Alvaro mentioned.

$arr = array('foo', 'bar', 'bash', 'monkey', 'badger');
foreach ($arr as $index => $key) {
    $$key = $data[$index];
}

or

$arr = array('foo', 'bar', 'bash', 'monkey', 'badger');
$result = array();
foreach ($arr as $index => $key) {
    $result[$key] = $data[$index];
}
extract($result);

In short, do not use "list", use arrays and associated arrays. Write helper functions to make your code clearer.

Answer 5

If you want to extract the elements of $data into the current context, you can do that with extract . You might find it useful to call array_intersect_key first to pare down $data to the elements that you want to extract.

Answer 6

May be try like this :

$arr = array('foo', 'bar', 'bash', 'monkey', 'badger');
$data = "$".implode(",", $arr);
echo str_replace(",", ",$", $data);