Question

我收到mysqli错误，它预计有2个参数，只有1个参数，但我看不出我错过了什么。

          // Create connection
          $con=mysqli_connect("connection stuff");

          // Check connection
          if (mysqli_connect_errno()) {
          echo "Failed to connect to MySQL: " . mysqli_connect_error();
          }

                    <?php
                        if(isset($_POST['submit1011']))
                        {
                        $query1 =mysqli_query ($con, "UPDATE blog_semicategory SET    english_navn_godkend=".$_POST['godkend_english101']." WHERE semikatagori_id=" . $id) or die(mysqli_error());
                        mysqli_query($query1);
                        echo('<META HTTP-EQUIV="refresh" CONTENT="1">'); 
                        echo "<h3>Oprettes...</h3>";
                        } 
                    ?>


                        <center><img src="../include/images/<?php echo $r1011->english_navn_godkend ?>.png"></center><br />

                        <center>
                            <form action="" method="post">
                            <input type="hidden" name="godkend_english101" value="2">
                            <input type="submit" name="submit1011" value="Godkend Navn" />
                            </form>
                        </center>

Answer 1

$query1 =mysqli_query ($con, "UPDATE blog_semicategory SET    english_navn_godkend=".$_POST['godkend_english101']." WHERE semikatagori_id=" . $id) or die(mysqli_error());
mysqli_query($query1);

$query1不是查询。请参阅mysqli_query的手册。该函数返回mysqli_result。实际上，您尝试执行两次相同的查询，第二次尝试不指定第一个链接参数。

Mysqli并期望2个参数，1给出

1 个答案: