mysqli_prepare()期望参数1为mysqli [英] mysqli_prepare() expects parameter 1 to be mysqli
本文介绍了mysqli_prepare()期望参数1为mysqli的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
在程序上使用 mysqli
功能如下:
db.php
<?php
define("DB_HOST", "host");
define("DB_LOGIN", "login");
define("DB_PASSWORD", "password");
define("DB_NAME", "myDb");
$link = mysqli_connect(DB_HOST, DB_LOGIN, DB_PASSWORD, DB_NAME);
if(!$link) {
echo 'ERROR: ' . mysqli_connect_errno() . ': ' . mysqli_connect_error();
}
?>
function.php
<?php
function addItemToCatalog($var1, $var2, $var3, $var4) {
$sql = 'INSERT INTO catalog (var1, var2, var3, var4)
VALUES (?, ?, ?, ?)';
if (!$stmt = mysqli_prepare($link, $sql)){
return false;
}
mysqli_stmt_bind_param($stmt, "ssii", $var1, $var2, $var3, $var4);
mysqli_stmt_execute($stmt);
mysqli_stmt_close($stmt);
return true;
}
?>
page.php
<?php
require_once ("db.php");
require_once ("function.php");
$var1 = $_POST['var1']; //showing without filtering methods
$var2 = $_POST['var2'];
$var3 = $_POST['var3'];
$var4 = $_POST['var4'];
if(!addItemToCatalog($var1, $var2, $var3, $var4)){
echo 'some error text';
}
else {
header("Location: success.php");
exit;
}
?>
使用此功能后
警告:mysqli_prepare()期望参数1为mysqli,在第5行的 function.php 中给出空值.
WARNING: mysqli_prepare() expects parameter 1 to be mysqli, null given in function.php on line 5.
有人吗?
推荐答案
两个文件db.php
和function.php
粘合在一起的方式我认为导致$link
被定义为全局变量-您需要使用 global
可以在函数中访问它:
The way the two files db.php
and function.php
are glued together I think results in $link
being defined as a global variable - you need to use global
to access it within function:
function addItemToCatalog($var1, $var2, $var3, $var4) {
global $link;
...
}
或通过参数将$link
明确赋予函数:
or give the $link
to the function explicitly through parameter:
function addItemToCatalog($var1, $var2, $var3, $var4, $link) {
...
}
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