this program must output a directory that you searched for how ever if its not found a message must appear that the org_name is not found,i don't know how to do that, i keep trying on some if-else but it just won't output it.
<?php
$con=mysql_connect("localhost","root","");
if(!$con) {
die('could not connect:'.mysql_error());
}
mysql_select_db("final?orgdocs",$con);
$org_name = $_POST["org_name"];
$position = $_POST["position"];
$result = mysql_query("SELECT * FROM directory WHERE org_name = '$org_name' OR position = '$position' ORDER BY org_name");
echo '<TABLE BORDER = "1">';
$result1 = $result;
echo '<TR>'.'<TD>'.'Name'.'</TD>'.'<TD>'.'Organization Name'.'</TD>'.'<TD>'.'Position'.'</TD>'.'<TD>'.'Cell Number'.'</TD>'.'<TD>'.'Email-Add'.'</TD>';
echo '</TR>';
while ( $row = mysql_fetch_array($result1) ){
echo '<TR>'.'<TD>'.$row['name'].'</TD>'.'<TD>'.$row['org_name'].'</TD>';
echo '<TD>'.$row['position'].'</TD>'.'<TD>'.$row['cell_num'].'</TD>'.'<TD>'.$row['email_add'].'</TD>';
echo '</TR>';
}
echo '</TABLE>';
?>
此程序条将输出搜索的目录,但如果找不到,则必须输出一条消息
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1条回答 默认 最新
- dongzang5815 2011-01-02 04:13关注
What you want is
mysql_num_rows
to check your query for how many result rows it has.http://de2.php.net/manual/en/function.mysql-num-rows.php
if (mysql_num_rows($result)>0) { // your thing above with mysql_fetch_array($result1) etc } else { echo 'nor found'; }
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