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Can I mix MySQL APIs in PHP? (4 answers)

Closed 7 years ago.

I got this error in my code and I don t know how to solve it my code:

<?php
session_start();
include_once"connect_to_mysql.php";

$db_host = "localhost";
// Place the username for the MySQL database here
$db_username = "root"; 
// Place the password for the MySQL database here
$db_pass = "****"; 
// Place the name for the MySQL database here
$db_name = "mrmagicadam";

// Run the actual connection here 
$myConnection= mysql_connect("$db_host","$db_username","$db_pass") or die ("could not connect to mysql");
mysql_select_db("mrmagicadam") or die ("no database");        
$sqlCommand="SELECT id, linklabel FROM pages ORDER BY pageorder ASC";
$query=mysqli_query($myConnection, $sqlCommand) or die(mysql_error());
$menuDisplay="";


while($row=mysql_fetch_array($query)) {
    $pid=$row["id"];
    $linklabel=$row["linklabel"];
$menuDisplay= <a href="index.php?pid=  .$pid .  ">  .$linklabel.  </a><br/> ;
}
mysqli_free_result($query);

?>

这是错误:

警告:我的sqli_query()预计参数1为我的sqli,在C:xampphtdocslimitlesslink_to_mysql.php第17条下提供的资源

What I am doing wrong?

Answer 1

http://php.net/manual/en/class.mysqli.php rel=“noretinger”>mysqli 和

你们需要使用

$myConnection= mysqli_connect("$db_host","$db_username","$db_pass") or die ("could not connect to mysql"); 

mysqli_select_db($myConnection, "mrmagicadam") or die ("no database");

mysqli在原有mysql展期上取得了许多改进,因此建议您使用mysqli。

Answer 2

你们正在使用不适当的yn子。如果你读到 mysqli_query() 你们会发现它需要两个参数。

混杂的我sqli_query(我sqli link , string$query [, int$resultmode = MYSQLI_STORE_RESULT]

mysql$link 一般而言,是指既定的我sqli与查询数据库链接的资源目标。

因此,解决这一问题有两个途径。

我sqli_query();

$myConnection= mysqli_connect("$db_host","$db_username","$db_pass", "mrmagicadam") or die ("could not connect to mysql"); 
$sqlCommand="SELECT id, linklabel FROM pages ORDER BY pageorder ASC";
$query=mysqli_query($myConnection, $sqlCommand) or die(mysqli_error($myConnection));

或,使用<代码>mysql_query() (现已过时)

$myConnection= mysql_connect("$db_host","$db_username","$db_pass") or die ("could not connect to mysql");
mysql_select_db("mrmagicadam") or die ("no database");        
$sqlCommand="SELECT id, linklabel FROM pages ORDER BY pageorder ASC";
$query=mysql_query($sqlCommand) or die(mysql_error());

正如在评论中指出的,人们意识到使用死亡只是犯错误。可能无意中向观众提供一些敏感信息。

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