How find Row Number of Value within Array?

Question

I've saved all my data to an array, and I want to get the 'name' of a supplied 'code'.

How to I get the array row of that code?

Also, is this the most efficient process?

 id  |  code  |  name     |  
__________________________
1   |  KNTY  |  Kentucky |
2   |  PURD  |  Purdue   |
3   |  TEXS  |  Texas    |


// Move data to array
$search = "SELECT * FROM table";
$query = mysqli_query($conn, $search);
while($row = mysqli_fetch_assoc($query)) {
   $array[] = $row;
}

// Code I want a name for
$code = "KNTY";
// MYSTERY STEP I NEED HELP WITH
$name = $array[$id]['name'];

Community · Accepted Answer · 2017-05-23 12:22:15Z

1

I edit with the hint of the comment of itachi. You can use the code as the key of the $array:

$search = "SELECT * FROM table";
$query = mysqli_query($conn, $search);

$array = array();

while($row = mysqli_fetch_assoc($query)) {
   $array[$row['code']] = $row['name'];
}

// Code I want a name for
$code = "KNTY";

$name = $array[$code];

edited May 23, 2017 at 12:22

CommunityBot

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answered Mar 17, 2015 at 4:10

Adrian Cid Almaguer

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3 Comments

James Patrick Over a year ago

Thanks Adrian - I'll be running this query a lot. I'd prefer not run multiple queries but rather get this information from the array.

James Patrick Over a year ago

Yes, I want to SELECT * from table only once and save to array. Then I want to manipulate the data within the array.

itachi Over a year ago

in the loop, use $array[$row['name']][] = $row;

Kevin · Accepted Answer · 2015-03-17 04:19:12Z

1

Yes you could do something like that, while inside the fetch loop, assign the code as key. This must be unique though:

$search = 'SELECT * FROM table_name';
$query = mysqli_query($conn, $search);
while($row = mysqli_fetch_assoc($query)) {
    // assign `$row['code']` as key to this rowset
    $array[$row['code']] = $row;
}

$code = 'KNTY';
if(isset($array[$code])) { // add some checking, you wouldn't want undefined index errors
    $name = $array[$code]['name'];
    echo $name;
} else {
    echo 'Sorry not found';
}

answered Mar 17, 2015 at 4:19

Kevin

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