PHP：MySQLの10行ごとに新しい行に画像を挿入するには？

Question

10行のデータが入力された後に、新しいランダム画像を1行（すべて単独で）挿入したいとします。

私はこのような入力何かすべきと思っています：（$カウンタ％10 == 0の場合

echo "<table border='1' CELLPADDING=5 STYLE='font-size:13px'>"; 
echo "<tr> <td><H3>No.</H3></td><td><H3>Date</H3></td>"; 
echo "<td><H3>First Name</H3></td> <td><H3>Last Name</H3></td>"; 
echo "<td><H3>City</H3></td><td><H3>Province</H3></td></tr>"; 

//declaring counter 
$count=0; 

// keeps getting the next row until there are no more to get 
while ($row = mysql_fetch_array($data, MYSQL_ASSOC)) { 
    //counter equals 
    $count=$count+1; 

    // Print out the contents of each row into a table 
    echo "</td><td>"; 
    echo $count; 
    echo "</td><td>"; 
    echo $row['DateIn']; 
    echo "</td><td>"; 
    echo $row['FirstName']; 
    echo "</td><td>"; 
    echo $row['LastName']; 
    echo "</td><td>"; 
    echo $row['City']; 
    echo "</td><td>"; 
    echo $row['Province_State']; 
    echo "</td></tr>"; 
} 
echo "</table>"; 

Answer 1

echo "<table border='1' CELLPADDING=5 STYLE='font-size:13px'>"; 
echo "<tr> <td><H3>No.</H3></td><td><H3>Date</H3></td><td><H3>First Name</H3></td> <td> <H3>Last Name</H3></td> <td><H3>City</H3></td><td><H3>Province</H3></td></tr>"; 

//declaring counter 
$count=0; 

// keeps getting the next row until there are no more to get 
while ($row = mysql_fetch_array($data, MYSQL_ASSOC)) { 

//counter equals 
$count++; 

//insert an image every 10 rows 
if($count==10){ 
    $count=0; 
    echo 'yourimage.jpg'; 
} 

// Print out the contents of each row into a table 
echo "</td><td>"; 
echo $count; 
echo "</td><td>"; 
echo $row['DateIn']; 
echo "</td><td>"; 
echo $row['FirstName']; 
echo "</td><td>"; 
echo $row['LastName']; 
echo "</td><td>"; 
echo $row['City']; 
echo "</td><td>"; 
echo $row['Province_State']; 
echo "</td></tr>"; 
} 
echo "</table>"; 

Answer 2

：

if($counter % 10 == 0) { 
    echo 'image file'; 
} 

しかし、私のコードでこれを組み込む方法は本当にわかりません）{ エコー '画像ファイル'; } 10、100ではありません:)。 $ count = $ count + 1ステートメントの後にこれを置いてください。

Answer 3

ここの助けのおかげでしかし、私はまだランダムな画像を表示するために取得することはできません（作品完成したコードですが、私は

echo "<table border='1' CELLPADDING=5 STYLE='font-size:13px'>"; 
    echo "<tr> <td><H3>No.</H3></td><td><H3>Date</H3></td><td><H3>First Name</H3></td> <td><H3>Last Name</H3></td> <td><H3>City</H3></td><td><H3>Province</H3></td></tr>"; 

    //declaring counter for data 
    $count=0; 

    //declaring counter for random image 
    $counter_for_image=0; 

    // keeps getting the next row until there are no more to get 
    while ($row = mysql_fetch_array($data, MYSQL_ASSOC)) { 

    //counter for actual count 
    $count=$count+1;  

    //counter for image count 
    //so I can reset count and 
    //not affect actual count 
    $counter_for_image++; 



    /* does not work 
    // start random image code 
    $images = array(
    0 => '1.jpg', 
    1 => '2.jpg', 
    ); 
    $image = $images[ rand(0,(count($images)-1)) ]; 
    $randomimage = "<img src=\"/http://wwww.my site.com/storm/images/".$image."\" alt=\"\" border=\"0\" />"; 
    //print($output); 
    // end random image code 
    */ 

    // start every 10th row image display code 
    if($counter_for_image==10){ 
    $counter_for_image=0; 

    echo '<tr><td colspan="6"><center><img src="http://www.mysite.com/storm/images/eric.jpg"/></center></td></tr>'; 

    /* would use this if I could get random image to work 
    echo '<tr><td colspan="6"><center>'; 
    print($randomimage); 
    echo '</center></td></tr>'; 
    */ 

    } // end every 10th row image display code 


    // Print out the contents of each row into a table 
    echo "</td><td>"; 
    echo $count; 
    echo "</td><td>"; 
    echo $row['DateIn']; 
    echo "</td><td>"; 
    echo $row['FirstName']; 
    echo "</td><td>"; 
    echo $row['LastName']; 
    echo "</td><td>"; 
    echo $row['City']; 
    echo "</td><td>"; 
    echo $row['Province_State']; 
    echo "</td></tr>"; 
} 
    echo "</table>";