php - Missing the result printed from database -
my purpose search book type input keyboard , count it. have error: counting number of book right when print, miss result. example, put 'f' show 7 result( right) print 2 two result. here code:
<form action="" method="post"> enter type of book here:<input type="text" size="20" name="sbt"> <br /> <input type="submit" name="sb" value="search"> </form> <table align="center" border="1" width="600"> <thead><tr align="center"> <tr align="center"> <td><b>book id</b></td> <td><b>book title</b></td> <td><b>book author</b></td> <td><b>pulished year</b></td> <td><b>book type</b></td> <td><b>status</b></td> </tr> <?php if (isset($_post['sb'])) { $s=""; if ($_post['sbt'] == null) { echo "please re-enter <br>"; } else { $s = $_post['sbt']; } $q = "select * book book_type '%$s%' "; $r= mysqli_query($conn,$q); while($row = mysqli_fetch_array($r)) { ?> <tr align="center"> <td><?php echo $row['book_no'];?></td> <td><?php echo $row['book_title'];?></td> <td><?php echo $row['book_author'];?></td> <td><?php echo $row['book_year'];?></td> <td><?php echo $row['book_type'];?></td> <td> <?php if ($row['book_quantity'] == 1) { echo "available"; } else { echo "not available"; } ?> </td> <?php $c=" select count(distinct(book_no)) totaltype book book_type '%$s%'"; $s= mysqli_query($conn, $c); if (mysqli_num_rows($s)> 0 ) { $row2= mysqli_fetch_array($s); echo " total book type result:"."{$row2['totaltype']}"; } } ?> <?php } ?> </table>
remove "%"
before $s
, try.
for example:
$query = mysql_query("select * table the_number '$yourphpvar%'");
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