在數組中搜索數據庫中的值

Question

我有一個數據庫，其中填充了員工ID和每個員工ID的對應員工名稱。 有沒有一種方法可以從數據庫中搜索員工ID的數組？ 我想Google並沒有幫助我，因為我不確定如何在搜索詞中表達。

我的想法是使用array_search（$ empID，$ currentArray）之類的東西。 然后遍歷數據庫中的每個員工ID，並將其與$ currentArray比較？ 我懷疑這是最有效的方法，但我仍在學習，因此我們將不勝感激。 如果有人願意花一些時間來幫助我，我可以發布更多需要的信息。 謝謝！

如果有人感興趣，請在此處編輯我的代碼：

 <?php 

//this variable tells us how many drupal nodes or 'paystub pages' we need to create
$nodeCount = 0;
$i = 0;

//needed for creating a drupal node
//for this code to work this script must be run from the root of the drupal installation
require_once './includes/bootstrap.inc';
drupal_bootstrap(DRUPAL_BOOTSTRAP_FULL);

if ($handle = opendir('/var/www/html/pay.mistequaygroup.com/upload')) 
{

    /* This is the correct way to loop over the directory. */
    while (false !== ($file = readdir($handle))) 
    {
       if ($file != "." && $file != "..") 
       {
            $nodeCount++;
            //We convert the pdf documents into text documents and move put them in the converted folder
            $command = "pdftotext /var/www/html/pay.mistequaygroup.com/upload/" . $file . " /var/www/html/pay.mistequaygroup.com/upload/converted/" . $file . ".txt";
            //Execute the command above
            $output = exec($command);

            //mark all the spots that TO THE ORDER OF shows up
            //echo array_search("TO THE ORDER OF", $currentArray);

            //echo $userName;


            //extract the employees name


            //print_r($currentArray);
            //echo '<pre>';
            //echo array_search("DATE AMOUNT", $currentArray);
            //echo '</pre>';

        }    
    }        
    closedir($handle);        
}

//subtract two because the folders "array" and "converted" are included because PHP does not differentiate
//between folders and files
$nodeCount = $nodeCount - 2; 

echo "<br />";
echo "I counted $nodeCount pdf files";
echo "<br />";

//open the directory
if ($handle2 = opendir('/var/www/html/pay.mistequaygroup.com/upload/converted')) 
{
    //check to see if we have reached the last file of our directory, if not stay in loop
    while (false !== ($currentText = readdir($handle2))) 
    {
        //filter out files named . and ..
       if ($currentText != "." && $currentText != "..") 
       {
               //Create a file for array to be printed to
               $createArray = fopen("/var/www/html/pay.mistequaygroup.com/upload/arrays/" . $currentText . ".txt", "w+") or die ("Cannot find file to create array, ID 2");


               //read the file we are on from the loop into the array 
               $currentArray = file("/var/www/html/pay.mistequaygroup.com/upload/converted/" . $currentText, FILE_SKIP_EMPTY_LINES) or die ("Cannot find file to create array, ID 1");

               //$countArray = array_search(". . . . . . . . . .", $currentArray);

               //echo $countArray;

               //print array to .txt file for debugging purposes
               $out = print_r($currentArray, true);
               fwrite($createArray, $out);
               fclose($createArray);


               //Loop?
            array_search($empID, $currentArray);

            //need to loop through the array we are on, looking for numbers that match the employee ID's 
            //OR we might have to search for names within a string of text and then get the corresponding ID for that user from the database? 

            //brainstorming 
               $query = SELECT * FROM `profile_values` WHERE `fid` = 2 AND `value` = $employeeID; 



               //DOES NOT WORK AS EXPECTED
               $indexEmpid = 0;
               foreach ($currentArray as $value)
               {
                   //set the value to 28 and it doubles each time the loop runs so there is no need to add 28 each time
                   //every 28th index in our array is an employee id
                   $indexEmpid = $indexEmpid + 28;
                   $currentEmployeeID = $currentArray[$indexEmpid];
                   echo "<br />";
                   echo "Employee ID's found: $currentEmployeeID";
                //echo "Employee ID's found: $currentArray[$indexEmpid]";
                echo "<br />";
                echo "IndexEmpid: $indexEmpid";            
               }




       }
     }

}


?>

Answer 1

是的，您可以使用SQL IN子句來做到這一點：

$empIDs = array(1,2,3);
$query = "SELECT name FROM emp WHERE ID IN (" . implode(',',$empIDs) . ")";

Answer 2

使用： where employeeID in (id list......) （用逗號分隔）

更多信息： 使用WHERE子句將數組傳遞給查詢