我在PHP中有一个简单的查询,但我无法让Like
工作。
以下是代码:
$var = $_GET['q'];
$trimmed = trim($var);
$query = "SELECT * FROM vm_regiony WHERE nazev LIKE "%$trimmed%" order by id LIMIT 10";
$result = mysql_query($query);
if(mysql_num_rows($result)==0){
echo "nothing";
echo "<br />";
echo $trimmed;
}else{
while($rene=mysql_fetch_array($result)){
$jmeno = $rene['nazev'];
echo '<a id="hled" onclick="javascript:vybrat()">'.$jmeno.'</a>';
答案 0 :(得分:5)
如果您需要使用单引号
$query = "SELECT * FROM vm_regiony WHERE nazev LIKE '%$trimmed%' order by id LIMIT 10";
答案 1 :(得分:0)
$query = "SELECT * FROM vm_regiony
WHERE nazev LIKE '%' . $trimmed . '%'
ORDER BY id LIMIT 10";