Question

我正在尝试使用以下查询从一个表中提取数据，但我需要它来运行第一个查询，如果得到结果则停止，然后如果没有，则运行第二个查询。每次运行搜索时，我都会得到两个结果，但是两个结果之一却在回声没有结果。我还计划添加更多查询，以便可以根据需要返回结果。我已经在这个问题上工作了几天，最终我想寻求帮助。预先感谢。

$conn = new mysqli("localhost", "user", "password", "dbname");                                                                               
if(!$conn) {                                                                                                                                              
    die("Connection failed: " . mysqli_connect_error());                                                                                                  
}                                                                                                                                                         

//echo $conn->host_info . "\n";                                                                                                                           

$sql = "SELECT * FROM SCANMSTR WHERE BadgeNumber LIKE $query AND Active = 1 AND ExpirationDate >= curdate()";                                             
$result = mysqli_query($conn, $sql);                                                                                                                      

if (mysqli_num_rows($result) > 0) {                                                                                                                       
   while($row = mysqli_fetch_assoc($result)) {                                                                                                            
        echo "Badge Number: " .$row["BadgeNumber"]. "- Name: " . $row["BadgeName"]. "<br>";                                                               
     }                                                                                                                                                    
}                                                                                                                                                         

else  {                                                                                                                                                   
  echo "Not in system1";                                                                                                                                  
}                                                                                                                                                         


$sql1 = "SELECT * FROM SCANMSTR WHERE BadgeNumber LIKE $query and Active = 0 AND Barred = 1 AND Lost = 1";                                                
$result1 = mysqli_query($conn, $sql1);                                                                                                                    

if (mysqli_num_rows($result1) >0) {                                                                                                                       
   while($row1 = mysqli_fetch_assoc($result1)) {                                                                                                          
        echo "Card not active";  }                                                                                                                        
      }                                                                                                                                                   

else {                                                                                                                                                    
   echo "Not in system";

//新添加的代码

$ row = mysqli_fetch_assoc（$ result）;
如果（$ row）{
  if（$ row ['Active']）{
        回显“徽章编号：”。$ row [“ BadgeNumber”]。 “-名称：”。$ row [“ BadgeName”]。 “
”;
     }
其他{
        回显“卡无效”；
}
如果（$ row）{
    if（$ row ['Barred']）
        回声“持卡人被禁止”；
}
如果（$ row）{
    if（$ row ['Lost']）
        回显“卡已报告丢失”；
}
      }
其他{
    回显“不在系统中”；

Answer 1

您将第二条语句放入第一条语句的第一条ELSE中，如下所示：

$conn = new mysqli("localhost", "user", "password", "dbname");
if(!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}

//echo $conn->host_info . "\n";

$sql = "SELECT * FROM SCANMSTR WHERE BadgeNumber LIKE $query AND Active = 1 AND ExpirationDate >= curdate()";
$result = mysqli_query($conn, $sql);

if (mysqli_num_rows($result) > 0) {
   while($row = mysqli_fetch_assoc($result)) {
        echo "Badge Number: " .$row["BadgeNumber"]. "- Name: " . $row["BadgeName"]. "<br>";
     }
}

else  {                                                                                                                                                   
$sql1 = "SELECT * FROM SCANMSTR WHERE BadgeNumber LIKE $query and Active = 0 AND Barred = 1 AND Lost = 1";
$result1 = mysqli_query($conn, $sql1);

if (mysqli_num_rows($result1) >0) {
   while($row1 = mysqli_fetch_assoc($result1)) {
        echo "Card not active";  }
      }

else {
   echo "Not in system";}

 }

Answer 2

您也许可以将两个查询都放在一个查询中，而只获取找到的第一行？像这样：

$sql = "SELECT BadgeNumber, BadgeName, Active FROM SCANMSTR WHERE BadgeNumber LIKE $query AND Active = 1 AND ExpirationDate >= CURDATE()
        UNION
        SELECT BadgeNumber, BadgeName, Active FROM SCANMSTR WHERE BadgeNumber LIKE $query and Active = 0 AND Barred = 1 AND Lost = 1
        LIMIT 1";

$result = mysqli_query($conn, $sql);
$row = mysqli_fetch_assoc($result);
if ($row) {
    if ($row['Active']) {
        echo "Badge Number: " .$row["BadgeNumber"]. "- Name: " . $row["BadgeName"]. "<br>";
    }
    else {
        echo "Card not active";  
    }
}
else {
    echo "Not in system";
}

编辑要将其扩展到N个支票，您可以添加更多

UNION
SELECT ...

$ sql的行，例如

$sql = "SELECT BadgeNumber, BadgeName, Active FROM SCANMSTR WHERE BadgeNumber LIKE $query AND Active = 1 AND ExpirationDate >= CURDATE()
        UNION
        SELECT BadgeNumber, BadgeName, Active FROM SCANMSTR WHERE BadgeNumber LIKE $query and Active = 0 AND Barred = 1 AND Lost = 1
        UNION
        SELECT ...
        UNION
        SELECT ...
        ...
        LIMIT 1";

由于LIMIT 1，查询将根据需要检查尽可能多的SELECT查询，以返回一行。因此，即使有多个SELECT匹配该人（不确定是否/应该吗？），您也会得到第一个，而不是全部。

或者，您可以简单地获取 all 检查所需的列，然后在后端逻辑中进行操作：

$sql = "SELECT BadgeNumber, BadgeName, Active, ExpirationDate 
        FROM SCANMSTR 
        WHERE BadgeNumber LIKE $query
        LIMIT 1";

$result = mysqli_query($conn, $sql);
$row = mysqli_fetch_assoc($result);
if ($row) {
    if ($row['Active']) {
        echo "Badge Number: " .$row["BadgeNumber"]. "- Name: " . $row["BadgeName"]. "<br>";
    }
    else if ($row['ExpirationDate'] < date('Y-m-d')) {
        echo "Card not active";
    }
}
else {
    echo "Not in system";
}

我个人认为我会这样做，感觉更简洁，重复SQL更少，等等。尽管这两种方法都可以工作，但：）..

我如何运行这些查询，以便一旦找到结果，它们就会停止

2 个答案: