在PHP数组中将SQLserv结果显示为文本
我有一个查询,我已经通过如下数组从Sql查询中提取了结果在PHP数组中将SQLserv结果显示为文本,php,arrays,sql-server,Php,Arrays,Sql Server,我有一个查询,我已经通过如下数组从Sql查询中提取了结果 <?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'
<?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 );
?>
$行
从数据库中取回一个数字。但是,我想显示$status数组中的文本
我尝试了几种方法,但最终返回一个空白屏幕
有什么建议吗?谢谢你的帮助 更改while循环,如下所示。检查状态号在
$status
中是否可用,如果可用,则显示其值,否则显示空白
while($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>";
}
如下更改您的while循环。检查状态号在
$status
中是否可用,如果可用,则显示其值,否则显示空白
while($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>";
}
请阅读以下内容:
请阅读以下内容:
是由PHP错误引起的空白屏幕吗?你应该检查你的日志。 因为在您的示例中,您编写的数组如下所示:
$status=(
'1' => "Request For Authorization",
'3' => "Pending",
'4' => "Scheduled For Review",
'5' => "Scheduled For Approval",
'8' => "Pending",
);
但您必须使用括号:
$status=[
'1' => "Request For Authorization",
'3' => "Pending",
'4' => "Scheduled For Review",
'5' => "Scheduled For Approval",
'8' => "Pending",
];
然后,您必须通过如下方式显示标签:
echo $status[$row['STATUS']];
顺便说一句,您应该使用(微)框架进行开发,因为您不应该(或必须)在数据库中编写查询并在同一文件中显示结果
Hth.
是由PHP错误引起的空白屏幕吗?你应该检查你的日志。 因为在您的示例中,您编写的数组如下所示:
$status=(
'1' => "Request For Authorization",
'3' => "Pending",
'4' => "Scheduled For Review",
'5' => "Scheduled For Approval",
'8' => "Pending",
);
但您必须使用括号:
$status=[
'1' => "Request For Authorization",
'3' => "Pending",
'4' => "Scheduled For Review",
'5' => "Scheduled For Approval",
'8' => "Pending",
];
然后,您必须通过如下方式显示标签:
echo $status[$row['STATUS']];
顺便说一句,您应该使用(微)框架进行开发,因为您不应该(或必须)在数据库中编写查询并在同一文件中显示结果
嗯