php/mysql elseif挑战

我正在尝试两个不同的查询，它们各自都可以正常工作，但我希望出现这样的情况，即其中一个会出现在PHP的elseif语句中，或者有人有更好的解决方案。下面是第一个和第二个语句的代码

第一句话

$retvval = mysql_query( $sqql);
$sqql = "SELECT `autoID`, `Name of Doctor`, `Email`, `Hospital of Practise`, `State`,
`Specialization`, `Professional Organization`, `webpage` FROM `doctor_locator`
WHERE state = '$states' AND `name of doctor` LIKE 'Dr%'";
$retvval = mysql_query( $sqql);
if(! $retvval )
{
  die('Could not get data: ' . mysql_error());
}

while($row = mysql_fetch_array($retvval))
{
 echo "{$row['autoID']}.  "."Name of Doctor: <a href = \"{$row['webpage']}\">{$row['Name of Doctor']} </a><br>"
    ."Email: {$row['Email']} <br>". "Hospital of Practise: {$row['Hospital of Practise']}
    <br>". "State:  {$row['State']} <br>" . "Specialization: {$row['Specialization']} <br> ";
}

$retval = mysql_query( $sql);
$sql = "SELECT `autoID`, `Name of Doctor`, `Email`, `Hospital of Practise`, `State`,
`Specialization`, `Professional Organization`, `webpage` FROM `doctor_locator`
WHERE state = '$states' AND specialization = '$specialize' ";
 $retval = mysql_query( $sql);
 if(! $retval )
 {
  die('Could not get data: ' . mysql_error());
 }

 while($row = mysql_fetch_array($retval))
 {
echo "{$row['autoID']}.  "."Name of Doctor: <a href = \"{$row['webpage']}">{$row['Name of Doctor']} </a><br>"."Email: {$row['Email']} <br>". "Hospital of Practise:{$row['Hospital of Practise']}<br>". "State:  {$row['State']} <br>" . "Specialization: {$row['Specialization']} <br> ";
}

$retvval=mysql\u查询（$sqql）；
$sqql=“选择'autoID'，'Name of Doctor'，'Email'，'Hospital of practice'，'State`，
`专业化“，”专业组织“，”医生定位器“网页”`
其中state='$states'和'name of doctor'类似于'Dr%'”；
$retvval=mysql\u查询（$sqql）；
如果（！$retvval）
{
die（'无法获取数据：'.mysql_error（））；
}
while（$row=mysql\u fetch\u数组（$retvval））
{
echo“{$row['autoID']}.”医生姓名：
“
“电子邮件：{$row['Email']}
”“执业医院：{$row['Hospital of practice']}”

““状态：{$row['State']}
”“专门化：{$row['Specialization']}
”；
}

第二项声明

$retvval = mysql_query( $sqql);
$sqql = "SELECT `autoID`, `Name of Doctor`, `Email`, `Hospital of Practise`, `State`,
`Specialization`, `Professional Organization`, `webpage` FROM `doctor_locator`
WHERE state = '$states' AND `name of doctor` LIKE 'Dr%'";
$retvval = mysql_query( $sqql);
if(! $retvval )
{
  die('Could not get data: ' . mysql_error());
}

while($row = mysql_fetch_array($retvval))
{
 echo "{$row['autoID']}.  "."Name of Doctor: <a href = \"{$row['webpage']}\">{$row['Name of Doctor']} </a><br>"
    ."Email: {$row['Email']} <br>". "Hospital of Practise: {$row['Hospital of Practise']}
    <br>". "State:  {$row['State']} <br>" . "Specialization: {$row['Specialization']} <br> ";
}

$retval = mysql_query( $sql);
$sql = "SELECT `autoID`, `Name of Doctor`, `Email`, `Hospital of Practise`, `State`,
`Specialization`, `Professional Organization`, `webpage` FROM `doctor_locator`
WHERE state = '$states' AND specialization = '$specialize' ";
 $retval = mysql_query( $sql);
 if(! $retval )
 {
  die('Could not get data: ' . mysql_error());
 }

 while($row = mysql_fetch_array($retval))
 {
echo "{$row['autoID']}.  "."Name of Doctor: <a href = \"{$row['webpage']}">{$row['Name of Doctor']} </a><br>"."Email: {$row['Email']} <br>". "Hospital of Practise:{$row['Hospital of Practise']}<br>". "State:  {$row['State']} <br>" . "Specialization: {$row['Specialization']} <br> ";
}

$retval=mysql\u查询（$sql）；
$sql=“选择'autoID'，'Name of Doctor'，'Email'，'Hospital of practice'，'State`，
`专业化“，”专业组织“，”医生定位器“网页”`
其中state='$states'和specialization='$specialize'；
$retval=mysql\u查询（$sql）；
如果（！$retval）
{
die（'无法获取数据：'.mysql_error（））；
}
while（$row=mysql\u fetch\u数组（$retval））
{
echo“{$row['autoID']}.”医生姓名：
““电子邮件：{$row['Email']}
”“执业医院：{$row['Hospital']}
““州：{$row['State']}
”，“专科：{$row['Specialization'}
”；
}

您可以这样做

$sql = "SELECT `autoID`, `Name of Doctor`, `Email`, `Hospital of Practise`, `State`,
`Specialization`, `Professional Organization`, `webpage` FROM `doctor_locator`
WHERE state = '$states' AND specialization = '$specialize' ";

$sqql = "SELECT `autoID`, `Name of Doctor`, `Email`, `Hospital of Practise`, `State`,
`Specialization`, `Professional Organization`, `webpage` FROM `doctor_locator`
WHERE state = '$states' AND `name of doctor` LIKE 'Dr%'";

$retval = (mysql_query($sqql) ? mysql_query($sqql) : mysql_query($sql));

if($retval){
      while loop
} 
else { 
      die(); 
}

---

这可以通过PDO实现：

/* Create new connection http://php.net/manual/en/pdo.construct.php */
$pdo = new Pdo($connectionString, $user, $pass);
/* Prepare the first query */
$statement = $pdo->prepare($sqql);
/* Without actually executing it, check if there is any error */
if ($statement->errorCode()) {
  /* If so, then prepare the second statement */
  $statement = $pdo->prepare($sql);
}

/* Execute it. It will be the first, if it was valid or the second, if the first was not valid. */
$statement->execute();

---

我不知道你到底想让它做什么？是否希望第二条语句仅在第一条语句失败时运行？您应该明确地修改代码缩进。创造奇迹！代码中没有elseif语句。你的意思是

如果（！$retvval）

？只需在SQL语句的末尾输入

xxx

，即可使其失败。使用

mysql\u num\u rows

检查查询中有多少个结果。但是请注意，

mysql\uuu

多年来一直被弃用，不再受支持。使用PDO或MySQLi。是的，我希望能够搜索其中一个information@h2oooooooo，这就是我需要您为我提供解决方案的原因。@anabisystems不是2个if语句，而是在一行if-else语句中生成$retval。