PHP Mysql替换while循环中的值
如何在查看期间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“type”数据是1,但在查看过程中,我需要替换为字符串值“home loan” 这是我用php编写的代码:PHP Mysql替换while循环中的值,php,mysql,Php,Mysql,如何在查看期间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“type”数据是1,但在查看过程中,我需要替换为字符串值“home loan” 这是我用php编写的代码: $loans = mysql_query("SELECT * FROM loans"); echo "<table cellspacing='2'>"; echo "<tr><th>ID</th><th>Name</th><th
$loans = mysql_query("SELECT * FROM loans");
echo "<table cellspacing='2'>";
echo "<tr><th>ID</th><th>Name</th><th width=70>Type</th><th width=70>Amount</th><th width=70>Duration</th><th>Installment</th><th></th><th></th></tr>";
?>
<form method="post" action="">
<?php
while ($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>";
$loans=mysql\u查询(“从贷款中选择*);
回声“;
回显“IDNameTypeAmountDurationInstallation”;
?>
试试这个
if($row["type"]==1){
echo "<td> Home Loan </td>";
} else {
echo "<td>other type</td>";
}
如果($row[“type”]==1){
呼应“住房贷款”;
}否则{
呼应“其他类型”;
}
希望它能有所帮助试试这个
if($row["type"]==1){
echo "<td> Home Loan </td>";
} else {
echo "<td>other type</td>";
}
如果($row[“type”]==1){
呼应“住房贷款”;
}否则{
呼应“其他类型”;
}
希望这将有助于一个简单的if条件将适用于此
if($row['type'] == 1) {
echo "<td>home loan</td>";
} else {
echo "<td>" . $row["type"] . "</td>";
}
if($row['type']==1){
呼应“住房贷款”;
}否则{
回显“$行[”类型“]”;
}
一个简单的if条件将适用于此
if($row['type'] == 1) {
echo "<td>home loan</td>";
} else {
echo "<td>" . $row["type"] . "</td>";
}
if($row['type']==1){
呼应“住房贷款”;
}否则{
回显“$行[”类型“]”;
}
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>";
$types=array(1=>“房屋贷款”);
$echo_type=isset($types[$row['type']])?$types[$row['type']]:$row['type'];
echo“$echo_type.”;
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>";
$types=array(1=>“房屋贷款”);
$echo_type=isset($types[$row['type']])?$types[$row['type']]:$row['type'];
echo“$echo_type.”;
while($row=mysql\u fetch\u数组($loans)){
printf(“%s%s%s%s%s%s”、$row[“贷款id”]、$row[“名称”]、((int)$row[“类型”]==1?“住房贷款”:“其他”)、$row[“金额”]、$row[“期限”]、$row[“分期付款”];
}
while($row=mysql\u fetch\u数组($loans)){
printf(“%s%s%s%s%s%s”、$row[“贷款id”]、$row[“名称”]、((int)$row[“类型”]==1?“住房贷款”:“其他”)、$row[“金额”]、$row[“期限”]、$row[“分期付款”];
}
您应该创建另一个表,其中包含与每种类型关联的值,并将该表上的值联接起来
表1贷款类型
id | loan_type
1 | Home Loan
2 | Other Loan
那你的问题呢
<?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>
身份证件
名称
类型
数量
期间
分期付款
您应该创建另一个表,其中包含与每种类型关联的值,并将该表上的值联接起来
表1贷款类型
id | loan_type
1 | Home Loan
2 | Other Loan
那你的问题呢
<?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>
身份证件
名称
类型
数量
期间
分期付款