我在标题中提到了警告，我的代码在这里:警告:mysqli_select_db()期望参数1是mysqli，in fim

Question

我在标题中提到了警告，我的代码在这里：警告：mysqli_select_db（）期望参数1为mysqli，字符串，在/cdc/slider.php在第3行

slider.php

<?php require_once('db/dbcon.php'); ?>
<?php mysqli_select_db($database_dbcon, $dbcon);
$query_slide ="SELECT * FROM photos_slide_tb";
$slide  = mysqli_query($query_slide , $dbcon) or die(mysqli_error());
$row_slide = mysqli_fetch_assoc($slide );
$totalRows_slide  = mysqli_num_rows($slide ); ?>
    <div id="sliderFrame">
        <div id="slider">
        <?php if ($totalRows_slide == 0) { ?><h1>&nbsp;</h1>
            <h1 align="center">Not available</h1>
        <?php } else do { ?> 
            <img src="photos/<?php echo $row_slide['location'];?>"  alt="<?php echo $row_slide['caption'];?>" />
        <?php } while ($row_slide = mysqli_fetch_assoc($slide)); ?>
        </div>
        <div id="htmlcaption" style="display: none;">
        </div>
    </div>

---

dbcon.php

<?php
$hostname_dbcon = "localhost";
$database_dbcon = "123";
$username_dbcon = "123";
$password_dbcon = "123";
$dbcon = mysqli_connect($hostname_dbcon, $username_dbcon, $password_dbcon) or trigger_error(mysqli_error(), E_USER_ERROR); 
?>

Answer 1

正确的参数顺序为 $connection, $db_name：

mysqli_select_db($dbcon, $database_dbcon);

Answer 2

您已将参数放到mysqli_select_db（）错误。

。

更正的是：

mysqli_select_db($dbcon, $database_dbcon);

参考：http：//php.net/manual/en/mysqli.select-db.php

Answer 3

在两个函数上正确参数

mysqli_select_db($database_dbcon, $dbcon);

应该是

mysqli_select_db($dbcon,$database_dbcon);

和另一个

mysqli_query($query_slide , $dbcon)

应该是

mysqli_query($dbcon, $query_slide)

Answer 4

传递错误的参数。第一个参数是connection object＆amp;第二个参数是database名称。

mysqli_select_db($dbcon, "Datebase Name");

对mysqli_query()

做同样的事情

mysqli_query($dbcon, $query_slide);