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我有一个更新页面，我可以在其中检查员工的头衔，无论他是医生还是护士。如果员工是医生/护士，则会显示 HTML 表单，如果不是医生/护士，则只会显示患者信息，无法编辑。但是我的代码以某种方式跳过了我想显示表单的部分，即使我以医生/护士的身份登录也是如此。你能帮我解决这个问题吗......

<?php
$a=$_SESSION['employeeID'];
$title="SELECT title FROM employee WHERE employeeID = '$a'";

if($title == 'nurse' OR $title == 'doctor')
{

echo '<form method="post" id="customForm" action="add_assessment.php">
<table>
<input type="hidden" name="res_id" value="' . $_GET['res_id'] . '" />
<tr>
<td><label for="name"><font style="color:white">Symptoms</font><font style="color:gray"></font></label>
<input id="name" name="symptoms" type="text" /></td>
<td><label for="name"><font style="color:white">Respiratory Rate</font></label>
<input id="name" name="respiratoryRate" type="text" /></td>
<td><label for="name"><font style="color:white">Temperature</font> <font     style="color:gray"></font></label>
<input id="name" name="temperature" type="text" /></td>
</tr>
<tr>
<td><label for="name"><font style="color:white">Blood Pressure</font></label>
<input id="name" input name="bloodPressure" type="text" class="input2"/></td>
<td><label for="name"><font style="color:white">Pulse Rate</font></label>
<input id="name" input name="pulseRate" type="text" /></td>
</tr>
<tr>
<td><label for="name"><font style="color:white">Chief Complaint</font></label>
<input id="name" input name="complaint" type="text" class="input2"/></td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td><input id="send" name="send" type="submit" value="Submit" /></td>
</tr>
</table>
</form>';
}
else 
{
$host="localhost"; 
$username="root"; 
$password=""; // password
$db_name="rhu"; // Database name
$tbl_name="assessment"; // Table name 
mysql_connect("$host", "$username", "$password");
mysql_select_db("$db_name");
$id = $_GET['res_id'];
$sql = mysql_query("SELECT * FROM assessment WHERE patientID='$id'");
while($row = mysql_fetch_array($sql))
{
echo "<p>ID: ".$id."</p>";
echo "<p>Assessment ID: ".$row['assessmentID']."</p>";
echo "<p>Symptoms: ".$row['symptoms']."</p>";
echo "<p>Respiratory Rate: ".$row['respiratoryRate']."</p>";
echo "<p>Temperature: ".$row['temperature']."</p>";
echo "<p>Blood Pressure: ".$row['bloodPressure']."</p>";
echo "<p>Pulse Rate: ".$row['pulseRate']."</p>";
echo "<p>Complaints: ".$row['complaint']."</p>";
echo "<p>Date: ".$row['date']."</p>";
echo "<br>";

}
}
?>
4

2 回答 2

1

在其当前状态下，您只需将字符串分配给变量$title。您实际上是在说那$title是字符串，"SELECT title FROM employee WHERE employeeID = '$a'";因此它正在跳过if($title == 'nurse' OR $title == 'doctor').

你也没有执行 MySQL 查询，先试试这个

$a=$_SESSION['employeeID'];
$sql = mysql_query("SELECT title FROM employee WHERE employeeID = '"$a"'");
while($row = mysql_fetch_array($sql)){
  $title = $row['title'];
if($title == 'nurse' OR $title == 'doctor')
{
   echo....

您还应该注意，mysql_*它已被弃用，并将在未来作为解决方案逐步淘汰 PHP。为了使您的代码适应未来，请考虑使用mysqli或 PDO 事务。

于 2013-01-06T03:34:37.230 回答
0

你可以这样试试—— 

$sql = mysql_query("SELECT title FROM employee WHERE employeeID = '$a'");
while($row = mysql_fetch_array($sql)){
$title = $row['title'];

if($title == 'nurse' OR $title == 'doctor')
{

...继续编码

于 2013-01-06T03:30:37.097 回答