Erhalten eines Arrays zu String-Konvertierung Fehler

Question

Kann mir jemand helfen, was ist das Problem mit diesem Code? Ich versuche, die abgerufenen Daten in einem Array zu speichern, und ich möchte basierend auf den Werten dieses Arrays. Ich bekomme einen Fehler von Array to string conversion. Der Datentypwert eines Arrays ist string

Hier ist der Code.

  $sql3 ="SELECT DISTINCT subj_descr FROM subj_enrolled WHERE enroll_ref = '$ref'"; 
      $results = mysqli_query($con, $sql3); 
      $data = array(); 
       while($row = mysqli_fetch_array($results)){ 
         $data[] = array($row['subj_descr']); 
       } 


      $sql ="SELECT * FROM notification WHERE subj_descr IN ({implode(',', $data})"; 
      $result = mysqli_query($con, $sql); 

      $count = mysqli_num_rows($result); 

Answer 1

entfernen array in Ihrem while-Schleife:

$sql3 ="SELECT DISTINCT subj_descr FROM subj_enrolled WHERE enroll_ref = '$ref'"; 
     $results = mysqli_query($con, $sql3); 
     $data = array(); 
      while($row = mysqli_fetch_array($results)){ 
        $data[] = $row['subj_descr']; 
      } 


     $sql ="SELECT * FROM notification WHERE subj_descr IN ({implode(',', $data})"; 
     $result = mysqli_query($con, $sql); 

     $count = mysqli_num_rows($result); 

Answer 2

Sie sind ein multidimensional Array erstellen und so diese Aussage ändern

$data[] = array($row['subj_descr']); 

zu

$data[] = $row['subj_descr']; 

Wie SQL IN Anweisung verwendet immer ein eindimensionales Array, so dass auch Änderung in Frage where-Klausel.

Ich habe alle geändert hat, versuchen Sie bitte die folgenden Code:

<?php 
    $sql3 ="SELECT DISTINCT subj_descr FROM subj_enrolled WHERE enroll_ref = '$ref'"; 
    $results = mysqli_query($con, $sql3); 
    $data = array(); 
    while($row = mysqli_fetch_array($results)){ 
     $data[] = $row['subj_descr']; 
    } 
    $dataStr = implode(',', $data); 

    $sql ="SELECT * FROM notification WHERE subj_descr IN (".$dataStr.")"; 
    $result = mysqli_query($con, $sql); 

    $count = mysqli_num_rows($result); 
    ?> 

Answer 3

eine neue Variable erstellen und darin implodieren.

dieses

$implodeAray = implode(",", $data); 
$sql ="SELECT * FROM notification WHERE subj_descr IN ($implodeAray)"; 

Answer 4

Sie das Element Ihres Arrays in einem anderen Array gespeichert Versuchen während Looping.

tun:

$sql3 ='SELECT DISTINCT subj_descr 
     FROM subj_enrolled 
     WHERE enroll_ref = "$ref"'; 

$results = mysqli_query($con, $sql3); 
$data = array(); 

while($row = mysqli_fetch_array($results)){ 
    //Your error was here 
    //Each elements is escaped for security reasons 
    $data[] = mysqli_escape_string($con,$row['subj_descr']);   
} 

//This implodes and puts a single quote around each element 
$dataIn= '\'' . implode('\', \'', $data) . '\''; 

$sql ="SELECT * FROM notification 
     WHERE subj_descr IN ($dataIn)"; 

$result = mysqli_query($con, $sql); 

$count = mysqli_num_rows($result);