PHP Mysql在while循环中替换值

Question

视图期间如何将数据库中的值替换为另一个值？

假设在数据库中，存储在数据库中的“类型”数据为1，但是在视图期间，我需要替换为字符串值“房屋贷款”

数据库中的样本数据

这是我在php中的编码：

$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>";

Answer 1

一个简单的if条件将对此起作用。

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

Answer 2

您应该创建另一个具有与每种类型关联的值的表，并将该值联接到该表上。

表loan_types

id | loan_type
1  | Home Loan
2  | Other Loan

然后你的查询

<?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>

Answer 3

尝试这个

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

希望对你有帮助

Answer 4

我喜欢这种方法：

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

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

Answer 5

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"]) ;
}