[英]PHP: How to Insert an Image in a new row for every 10th row in the MySQL?
我想在填充10行數據后,在一行中插入一個新的隨機圖像(全部為自己)。
我想我應該輸入這樣的東西:
if($counter % 10 == 0) {
echo 'image file';
}
但不確定如何在我的代碼中加入它:
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 $row['Province_State'];
echo "</td></tr>";
}
echo "</table>";
if($ counter%10 == 0){echo'image file'; 10,而不是100 :)。 把這個$ count = $ count + 1語句放在td colspan =你有多少列之后。
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>";
感謝幫助,這里有完整的代碼可以使用(但是,我仍然無法顯示隨機圖像,但我可以得到一個圖像:-)
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>";
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