我有一个表单,用于在名为t_id
的表中插入数据。它插入成功。但是当我尝试获取它if(isset($_POST['submit'])) {
$cat = $_POST['categories_chk'];
$tag = $_POST['tag'];
// echo print_r($cat);
$query1 = "select * from tags where tag_name='$tag' ";
$search_cat = mysqli_query($con, $query1);
$count = mysqli_num_rows($search_cat);
if ($count == 0) {
$query = "insert into tags values('','$tag')";
$result = mysqli_query($con, $query);
if ($result) {
$fetch_id = mysqli_query($con, "select t_id from tags where tag_name='$tag'");
while ($row = mysqli_fetch_array($fetch_id, MYSQLI_ASSOC)) {
$tag_id = $row['t_id'];
echo $tag_id;
}
}
}
else {
echo "<script>alert('Row with the same name exists! ')</script>";
} // This is line 130
}
时,输出来了
警告:mysqli_fetch_array()要求参数1为mysqli_result,第130行的\ tags.php中给出布尔值
代码
now.getDay()
答案 0 :(得分:0)
选中此项,您需要一个条件来检查true
if ($search_cat = mysqli_query($con, $query1))
{
$count = mysqli_num_rows($search_cat);
}
答案 1 :(得分:0)
您可以简单地使用last insert ID函数,该函数返回最新查询中使用的自动生成的ID
if ($result) {
$fetch_id = mysqli_insert_id($con);
echo $fetch_id;
}