Display SQLserv result as text in PHP array

Question

I have a query where I'm already pulling back the results from a Sql query via an array as below.

<?php

$sql = "SELECT inc.STATUS as STATUS,
FROM dbo.HELP_DESK as inc";

$result = sqlsrv_query($conn, $sql);

$status=(
'1' => "Request For Authorization",
'3' => "Pending",
'4' => "Scheduled For Review",
'5' => "Scheduled For Approval",
'8' => "Pending",
);

echo "<table><tr id=\"header\"><td><center>Status</center></td><tr>";

while($row = sqlsrv_fetch_array($result, SQLSRV_FETCH_ASSOC)) {

echo "<tr><td><b><center> " . $row["STATUS"] . "</center></b></td></tr>";
}

echo "</table>";

sqlsrv_close( $conn );
?>

The $row fetches back a number from the database. However, I'm wanting to display the text from my $status array.

I've tried several methods but end up with a blank screen returned.

Any suggestions? Appreciate the help.

B. Desai · Accepted Answer · 2017-11-17 11:32:40Z

2

change your while loop as below. check status number is available in $status or not if available then display its value otherwise display blank

while($row = sqlsrv_fetch_array($result, SQLSRV_FETCH_ASSOC)) {
    $status_str = isset($status[$row["STATUS"]])?$status[$row["STATUS"]]:"";
    echo "<tr><td><b><center> " . $status_str . "</center></b></td></tr>";
}

Also change your array to(You forgot array keyword)

$status=array(
'1' => "Request For Authorization",
'3' => "Pending",
'4' => "Scheduled For Review",
'5' => "Scheduled For Approval",
'8' => "Pending",
);

edited Nov 17, 2017 at 11:32

answered Nov 17, 2017 at 11:25

B. Desai

16.4k5 gold badges28 silver badges48 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Roman Kovalchuk · Accepted Answer · 2017-11-17 11:28:03Z

0

Read also about this: http://php.net/manual/en/function.strtr.php

strtr($row["STATUS"], $status);

answered Nov 17, 2017 at 11:28

Roman Kovalchuk

513 bronze badges

Comments

Fiber Optic · Accepted Answer · 2017-11-17 11:49:28Z

Is the the blank screen caused by a PHP error? You should check your logs. Because, in your example, your wrote the array like this:

    $status=(
'1' => "Request For Authorization",
'3' => "Pending",
'4' => "Scheduled For Review",
'5' => "Scheduled For Approval",
'8' => "Pending",
);

But you have to use brackets instead:

$status=[
'1' => "Request For Authorization",
'3' => "Pending",
'4' => "Scheduled For Review",
'5' => "Scheduled For Approval",
'8' => "Pending",
];

Then you will have to display the the label by something like this:

echo $status[$row['STATUS']];

BTW, you should use a (micro) framework to develop, because you should (or must) not write a query in your DB and display the resuslts in the same file.

HTH.

Collectives™ on Stack Overflow

Display SQLserv result as text in PHP array

3 Answers 3

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related