无法使用foreach循环删除数据库行-php / mysql

Question

I trying to delete some data from three tables,

1:- ailments 2:- jnctn_ailments_symptoms 3:- symptoms

I tried to delete the Link first from Junction Table, Then Delete The Ailment (vise versa too) and then final check the symptoms for any other link with another ailment, if not delete the symptom from symptom table. but I am failing at on step in foreach loop, I am checking the occurrences for symptoms beside the ailment deleted. everything is fine but when it comes to deleting the symptom I get mysql error " You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1".

I the ids for the symptoms are retrieved successfully but not delete query not working, somehow.

please see the following code and let me know where I am mistaken.

if(isset($_GET['del']))
{
    $delID = $_GET['del'];



    $originalSymptoms="";
    $symptomO[] = array();
    echo $delID;
    $symp = "SELECT symptomID  FROM symptoms 
            INNER JOIN jnctn_ailments_symptoms
            ON jnctn_ailments_symptoms.FK_symptomID= symptoms.symptomID
            INNER JOIN ailments
            ON ailments.ailmentID = jnctn_ailments_symptoms.FK_ailmentID
            WHERE ailments.ailmentID IN ('".$delID."')";


    mysql_select_db($dbName);
    $sympRes = mysql_query($symp,$con) or die(mysql_error());

        while($symprow = mysql_fetch_assoc($sympRes))
             {
               if(empty($originalSymptoms))
               {
                $originalSymptoms = $symprow['symptomID'];
               }
                else
                {
                $originalSymptoms = $originalSymptoms.",".$symprow['symptomID']; 
                }
              }




   echo $originalSymptoms;


     $newSymptoms = explode("," , $originalSymptoms);
    foreach($newSymptoms as $symptom)
    {
    $originalSymptomsArray[] = $symptom;

    }
       echo count($originalSymptomsArray);

        $delAilment = "DELETE FROM ailments WHERE ailmentID='$delID'";
    $delAilmentResult = mysql_query($delAilment,$con) or die(mysql_error());

   $delLink = "DELETE FROM  jnctn_ailments_symptoms WHERE FK_ailmentID ='$delID'";
   $delLinkResult = mysql_query($delLink,$con) or die(mysql_error());

    foreach($originalSymptomsArray as $symptom)
    {
        echo $symptom."<br>";




        $sql2 = "SELECT * FROM jnctn_ailments_symptoms WHERE FK_symptomID=".$symptom;
        $result2 = mysql_query($sql2,$con);
        $count = mysql_num_rows($result2);
        if(!$result2 || $result2 != 0)
        {
            echo mysql_error();
            $delSymptom = false;
        }
        else
        {
            $delSymptom = true;
        }


        if($delSymptom)
        {

            $sqlDel = "DELETE FROM symptoms WHERE symptomID ='$symptom'";
            $delResult = mysql_query($sqlDel,$con)or die(mysql_error());
            if(!$delResult)
            {
                echo mysql_error();
            }
            else
            {
                echo "Symptoms Deleted!";
            }
        }




    }

I have tried FK_symptomID='$symotom'"; , concatenation etc, nothing worked.

Answer 1

尝试将以下提到的所有查询连接起来，

$sqlDel = "DELETE FROM symptoms WHERE symptomID ='".$symptom."'";

Answer 2

somehow following code worked

 foreach($originalSymptomsArray as $symptom)
    {


        $sql2 = "SELECT `ID`, `FK_ailmentID`, `FK_symptomID` FROM `jnctn_ailments_symptoms` WHERE `FK_symptomID`='".$symptom."'";

        $result2 = mysql_query($sql2,$con);
        $count = mysql_num_rows($result2);
        if($count>0)
        {
            echo mysql_error();
            $delSymptom = 0;
        }
        else
        {

            $delSymptom = 1;
        }


        if($delSymptom==1)
        {

            $sqlDel = "DELETE FROM symptoms WHERE symptomID ='".$symptom."'";
            $delResult = mysql_query($sqlDel,$con)or die(mysql_error());
            if(!$delResult)
            {
                echo mysql_error();
            }

        }




    }

thank you for concatenation suggestion, and thank you everybody.