我在让我的查询排除我的“软删除”列时遇到问题 我已将我的代码附在下面关于让它工作的任何提示？它显示具有deleted= 1 的行，我只需要它来显示 0
$query=mysql_real_escape_string($_GET['query']);
$query_for_result=mysql_query("SELECT * FROM Assets WHERE
`Badge` like '%".$query."%' or
`Status` like '%".$query."%' or
`First Name` like '%".$query."%' or
`Last Name` like '%".$query."%' or
`Service Tag` like '%".$query."%' or
`Asset Tag` like '%".$query."%' and
`deleted` = '0' ")or die(mysql_error());
if(mysql_num_rows($query_for_result1)==0){
}
while($data_fetch=mysql_fetch_array($query_for_result))
{
echo '<h3>';
echo "<a href=\"modify.php?id=" . $data_fetch['id'] . "\">" . $data_fetch['First Name'] . ' ' . $data_fetch['Last Name'] . "</a>";
echo '<br/><b>Badge:</b> '. $data_fetch['Badge'];
echo '<br/> <b>Service Tag:</b> '. $data_fetch['Service Tag'];
echo ' <b>Asset Tag:</b> '. $data_fetch['Asset Tag'];
echo '<br/> <b>Status:</b> '. $data_fetch['Status'];
echo '<br/><b>Employee Status: </b>';
if( $data_fetch['Employee Status'] == 'Active' ){
echo '<font color="#32CD32">' . $data_fetch['Employee Status'] . '</font>';
}
elseif( $data_fetch['Employee Status'] == 'Terminated' ){
echo '<font color="red">' . $data_fetch['Employee Status'] . '</font>';}
echo '<br/> <b>Last Modified:</b> '. $data_fetch['Last Modified'];
echo "<span> </span>";
echo '</h3>';
echo '<br/><p>' ;
}