(菲律宾比索）如果，否则无法正确检索条件中的数据 - (PHP) If, Else not retrieve data in the condition properly

(PHP) If, Else not retrieve data in the condition properly

我这里有一个 If-Else 条件的问题。正如下面的代码，

if($status == 'New'){
    $resultx0 = mysql_query("SELECT * FROM application_data_file WHERE Position_ID = '".$pos_id."'  && Application_Status = 'New'  OR 'Received'  OR 'Checking' ");
}
elseif ($status == 'Not qualified'){
    $resultx0 = mysql_query("SELECT * FROM application_data_file WHERE Position_ID = '".$pos_id."'  && Application_Status = 'Failed'  OR 'Not qualified' ");
}
elseif ($status == 'No Condition'){
    $resultx0 = mysql_query("SELECT * FROM application_data_file WHERE Position_ID = '".$pos_id."' ");  
}

新"和"不合格"条件仅适用于"application_data_file"表中的第一个条件。"新"条件将仅检索"新"，但没有检索"已接收"或"正在检查"，因为我在条件中输入，"不合格"条件将仅检索"失败"而不检索"不合格"。

我对此一无所知。条件有什么问题吗？

干杯。

像这样尝试

mysql_query("SELECT * FROM application_data_file 
             WHERE Position_ID = '".$pos_id."' 
             AND Application_Status IN ('New','Received','Checking')");

您的 mysql 语法不正确

Application_Status = 'New' OR 'Received' OR 'Checking'

这应该是

Application_Status = 'New' OR Application_Status = 'Received' OR Application_Status = 'Checking'

或

Application_Status IN ('New','Received','Checking')

请仔细阅读代码。

在两种情况下，我们需要检查仓位 ID：

New和Not qualified .对于No Condition，没有检查。

因此，仅添加两个条件：New和Not qualified。

获取一个位置 ID 数组，如果条件添加IN子句，否则不执行任何操作。

// Initiate a blank array
$positionIds = array();
$status = 'New';
if ($status == 'New') {
    $positionIds[] = 'New';
    $positionIds[] = 'Received';
    $positionIds[] = 'Checking';
}
else if ($status == 'Not qualified') {
    $positionIds[] = 'Failed';
    $positionIds[] = 'Not qualified';
}
$sql = "SELECT * FROM application_data_file WHERE Position_ID = '".$pos_id."";
if (! empty($positionIds)) {
    $comma_separated = implode("','", $positionIds);
    $comma_separated = "'".$comma_separated."'";
    $sql .= " AND Application_Status IN ( " . $comma_separated .  ")";
}
$resultx0 = mysql_query($sql);

请不要使用 mysql_* 函数，因为它们已被弃用和删除由于安全原因，完全在 PHP 7 中。

SQL 语句格式不正确; 在您的情况下，我建议使用 IN 检查Application_Status字段(如下所示(;

$pos_id = mysql_real_escape_string($pos_id);
if($status == 'New') {
    $query = "SELECT * FROM application_data_file"
            . " WHERE Position_ID='" . $pos_id . "'"
            . " AND Application_Status IN ('New', 'Received', 'Checking')";
} elseif ($status == 'Not qualified') {
    $query = "SELECT * FROM application_data_file"
            . " WHERE Position_ID='" . $pos_id . "'"
            . " AND Application_Status IN ('Failed', 'Not qualified')";
} else {
    // Includes the case where $status == 'No Condition'
    $query = "SELECT * FROM application_data_file"
            . " WHERE Position_ID='" . $pos_id . "'";
}
$resultx0 = mysql_query($query);