PHP mysql在4列中顯示查詢

Question

如何在4列中顯示此查詢結果

<?php
$direction = $_POST['direction']; 
$sumword = $_POST['sumword'];  
$length = $_POST['length'];  
$con = mysql_connect("localhost","elsha","12q(5PSZ.");
$db = mysql_select_db("elsha",$con);
$query = "SELECT answer FROM words WHERE direction = '$direction' AND sumword = '$sumword' AND sumletter = '$length'";
$result = mysql_query($query);

if(mysql_error()) { 
  //check that no error has occurred first; take this out in production or make more graceful handling
  die(mysql_error());
}

if(mysql_num_rows($result) == 0) {
  echo "No Results";
} else { 
  while($row = mysql_fetch_array($result)) {
    echo '<img src="/images/'. $row['0'].'.jpg"><br> '. $row['0'].'<br>';
  }
}
?>

像這樣 ：

result1 result2 result3 result4

結果5結果6結果7結果8

Answer 1

嘗試這個

$i = 0;

while($row = mysql_fetch_array($result)) {
    echo '<img src="/images/'. $row['0'].'.jpg"> ';
    $i++;
    if($i % 4 == 1 && $i!=1){
        echo '<br>';
    }
}

Answer 2

使用像

$flag=0;
while($row = mysql_fetch_array($result)) {
     if(($flag%4)==0) {
       echo '<tr>';
     } 
     echo '<td><img src="/images/'. $row['0'].'.jpg"><br> '. $row['0'].'</td>';
    if(($flag%4)==3) {
      echo '</tr>';
      $flag=-1;
     } 
     $flag++;
}

Answer 3

用以下代碼替換您的一會兒。 嘗試這個

$a=1;
while($row = mysql_fetch_array($result)) {
    if($a%4==0) {
        echo '<img src="/images/'. $row['0'].'.jpg">'.    $row['0'].'<br>';
    } //4th line with break
    else {
        echo '<img src="/images/'. $row['0'].'.jpg"> '. $row['0'].' '; // prints first 3 lines 
    }
    $a=$a+1;
}

Answer 4

我認為您正在嘗試在圖像下方顯示圖像名稱。 嘗試這個：

$i = 1;
while($row = mysql_fetch_array($result)) {
       echo '<div class="imgs">';
       echo '<img src="/images/'. $row['0'].'.jpg"><br> '. $row['0'];
       echo '</div>';
       if ($i === 4) {
           echo '<div class="clear"></div>';
           $i = 1;
       } else {
           $i++;
       }
 }

然后設置div的樣式

.imgs {
  float: left;
  /* other styles */
}
.clear {
  clear: both;
}