大家好,我目前正处于想要以正确的方式将所有mysql_ *代码转换为mysqli_ *的情况,但我不知道该怎么做。所以我真的需要你的帮助和建议来帮助我的代码。
$sql = "SELECT * FROM studentprofile ";
$query = mysql_query($sql);
if (isset($_POST['search'])) {
$search_term = mysql_real_escape_string($_POST['search_box']);
$sql .= "WHERE name = '{$search_term}'";
$sql .= " OR class = '{$search_term}'";
}
数据输入:
Filter class/name : <input type="text" name="search_box" value="" />
要获取数据,请使用以下代码:
while ($row = mysql_fetch_array($query)) {
<td width="500"><?php echo $row['name']; ?></td>
代码在localhost上工作正常,但每当我将它用于网站主机时,它都需要在mysqli_ *中,我总是遇到这个问题:
“mysqli_real_escape_string()需要2个参数,1个给出...”。
感谢。
答案 0 :(得分:-1)
使用此代码:
$con=mysqli_connect("localhost","username","password","dbname");
$sql = "SELECT * FROM studentprofile ";
$query = mysqli_query($con, $sql);
if (isset($_POST['search'])) {
$search_term = mysqli_real_escape_string($con, $_POST['search_box']);
$sql .= "WHERE name = '{$search_term}'";
$sql .= " OR class = '{$search_term}'";
}
获取数据使用此代码:
<?php while ($row = mysqli_fetch_array($query,MYSQLI_ASSOC)) {?>
<td width="500"><?php echo $row['name']; ?></td>
<?php }?>