PHP Mysql Replace value in while loop

Question

How to replace value in database to another value during view?

Let say in database , 'type' data stored in database is 1 but during view i need to replace to string value "home loan"

This is my coding in php:

$loans = mysql_query("SELECT * FROM loans");

echo "<table cellspacing='2'>";

echo "<tr><th>ID</th><th>Name</th><th width=70>Type</th><th width=70>Amount</th><th width=70>Duration</th><th>Installment</th><th></th><th></th></tr>";
?>
<form method="post" action="">
<?php
while ($row = mysql_fetch_array($loans)) {

    echo "<tr>";

    echo "<td>" . $row["loan_id"] . "</td>";

    echo "<td>" . $row["name"] . "</td>";

    echo "<td>" . $row["type"] . "</td>";

    echo "<td>" . $row["amount"] . "</td>";

    echo "<td>"  . $row["duration"] .  "</td>";

    echo "<td>"  . $row["installment"] .  "</td>";

    echo "</tr>";

}

echo "</table>";

BigToach · Accepted Answer · 2013-08-02 04:56:28Z

You should create another table with values associated with each of the types and join the value on that table.

Table loan_types

id | loan_type
1  | Home Loan
2  | Other Loan

Then your query

<?php
$loans = mysql_query("SELECT loans.*, loan_types.`loan_type` FROM loans LEFT JOIN loan_types ON loans_types.`id` = loans.`type`");
?>

<table cellspacing='2'>
    <tr>
        <th>ID</th>
        <th>Name</th>
        <th width=70>Type</th>
        <th width=70>Amount</th>
        <th width=70>Duration</th>
        <th>Installment</th>
    </tr>
    <?php while ($row = mysql_fetch_array($loans)) { ?>
    <tr>
        <td><?php echo $row["loan_id"]; ?></td>
        <td><?php echo $row["name"]; ?></td>
        <td><?php echo $row["loan_type"]; ?></td>
        <td><?php echo $row["amount"]; ?></td>
        <td><?php echo $row["duration"]; ?></td>
        <td><?php echo $row["installment"]; ?></td>
    </tr>
    <?php } ?>
</table>

Sonu Sindhu · Accepted Answer · 2013-08-02 04:46:04Z

1

try this

  if($row["type"]==1){ 
     echo "<td> Home Loan </td>";
  } else { 
     echo "<td>other type</td>"; 
  }

Hope it will help

answered Aug 2, 2013 at 4:46

Sonu Sindhu

1,80216 silver badges27 bronze badges

Comments

Yogesh Suthar · Accepted Answer · 2013-08-02 04:46:31Z

1

A simple if condition will work for this.

if($row['type'] == 1) {
    echo "<td>home loan</td>";
} else {
    echo "<td>" . $row["type"] . "</td>";
}

answered Aug 2, 2013 at 4:46

Yogesh Suthar

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Comments

steven · Accepted Answer · 2013-08-02 04:48:48Z

0

i like this approach:

$types = array(1=>"Home Loan");

$echo_type = isset($types[$row['type']])?$types[$row['type']]:$row['type'];
echo "<td>".$echo_type."</td>";

answered Aug 2, 2013 at 4:48

steven

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Comments

Moak · Accepted Answer · 2013-08-02 04:49:56Z

0

while ($row = mysql_fetch_array($loans)) {

   printf("<tr><td>%s</td><td>%s</td><td>%s</td><td>%s</td><td>%s</td><td>%s</td></tr>",$row["loan_id"],$row["name"], ((int)$row["type"] === 1 ? 'home loan':'other'),$row["amount"], $row["duration"], $row["installment"]) ;
}

answered Aug 2, 2013 at 4:49

Moak

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