2017-06-15 66 views
0

我想從我的數據庫中的列回合得到「回合」的價值。所以,我想爲每個辯論/帖子獲得一個字段的值(1-5)。基於價值/數量,它應該顯示不同的東西。當我認爲它的值爲NULL時,即使該辯論/帖子的db字段值是4.這不僅僅是辯論,但是對於所有這些都會發生。如何獲得列中字段的實際值並將其分配給名爲$ rounds的變量。這個變量需要具有每次辯論的價值,而不僅僅是辯論。從每個字段的列中獲取值,但它返回NULL?

<?php 
$servername = ""; 
$username = ""; 
$password = ""; 
$dbname = ""; 

$con = new mysqli($servername, $username, $password, $dbname); 

if ($con->connect_error) { 
    die("Connection failed: " . $con->connect_error); 
} 
else{ 
    $sql = "SELECT rounds FROM vf_Discussion"; 
    $result = $con->query($sql); 

    $allRounds = $result->fetch_row(); 
    $rounds = $allRounds[0]; 

    var_dump($rounds); 

} 
    mysqli_close($con); 

     $rounds1 =  '<h2 class="CommentHeading">Round 1 (Pro)</h2> <br> <h2 class="CommentHeading">Round 1 (Con) </h2>'; 

     $rounds2 = '<h2 class="CommentHeading">Round 2 (Pro)</h2> <br> <h2 class="CommentHeading">Round 2 (Con)</h2>'; 

     $rounds3 = '<h2 class="CommentHeading">Round 3 (Pro)</h2> <br> <h2 class="CommentHeading">Round 3 (Con)</h2>'; 

     $rounds4 =  '<h2 class="CommentHeading">Round 4 (Pro)</h2> <br> <h2 class="CommentHeading">Round 4 (Con)</h2>'; 

     $rounds5 = '<h2 class="CommentHeading">Round 5 (Pro)</h2> <br> <h2 class="CommentHeading">Round 5 (Con)</h2>'; 

    foreach($allRounds as $rounds) { 
     if ($rounds == 1) { 
      echo $rounds1; 
      foreach ($Sender->Data('Answers') as $Row) { 
       $Sender->EventArguments['Comment'] = $Row; 
       WriteComment($Row, $Sender, Gdn::Session(), 0); 
      } 
     } 
     if ($rounds == 2) { 
      echo $rounds1; 
      echo $rounds2; 
      foreach ($Sender->Data('Answers') as $Row) { 
       $Sender->EventArguments['Comment'] = $Row; 
       WriteComment($Row, $Sender, Gdn::Session(), 0); 
      } 
     } 
     if ($rounds == 3) { 
      echo $rounds1; 
      echo $rounds2; 
      echo $rounds3; 
      foreach ($Sender->Data('Answers') as $Row) { 
       $Sender->EventArguments['Comment'] = $Row; 
       WriteComment($Row, $Sender, Gdn::Session(), 0); 
      } 
     } 
     if ($rounds == 4) { 
      echo $rounds1; 
      echo $rounds2; 
      echo $rounds3; 
      echo $rounds4; 
      foreach ($Sender->Data('Answers') as $Row) { 
       $Sender->EventArguments['Comment'] = $Row; 
       WriteComment($Row, $Sender, Gdn::Session(), 0); 
      } 
     } 
     if ($rounds == 5) { 
      echo $rounds1; 
      echo $rounds2; 
      echo $rounds3; 
      echo $rounds4; 
      echo $rounds5; 
      foreach ($Sender->Data('Answers') as $Row) { 
       $Sender->EventArguments['Comment'] = $Row; 
       WriteComment($Row, $Sender, Gdn::Session(), 0); 
     } 
    } 

    } 
     ?> 
+1

的可能的複製[正從列中的值,而是返回null?](https://stackoverflow.com/questions/44574090/getting-the-迭代value-from-column-but-returns-null) – Phil

回答

0

您應該通過對象

<?php 







    $servername = ""; 
    $username = ""; 
    $password = ""; 
    $dbname = ""; 


    $con = new mysqli($servername, $username, $password, $dbname); 

    if ($con->connect_error) { 
     die("Connection failed: " . $con->connect_error); 
    } 
    else{ 
     $sql = "SELECT rounds FROM vf_Discussion"; 
     $result = $con->query($sql); 


     while($row = mysql_fetch_array($result)) 
     { 

      var_dump($row); 
    } 




    } 
+0

不起作用,您能否提供其他工作解決方案? – julereca

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