PHP mysql按4列显示查询

Question

如何在4列中显示此查询结果

<?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>';
  }
}
?>

如下所示：

result1 result2 result3 result4

result5 result6 result7 result8

Answer 1

尝尝这个

$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

使用像这样的标志

$flag=0;
while($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

将您的while替换为以下代码。尝尝这个

$a=1;
while($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;
}