Php Mysqli,夹在;还是死;及;sqlierror";
我在开始治疗时遇到了麻烦。以下是我的php:Php Mysqli,夹在;还是死;及;sqlierror";,php,mysqli,dreamweaver,wampserver,Php,Mysqli,Dreamweaver,Wampserver,我在开始治疗时遇到了麻烦。以下是我的php: <?php session_start(); $con=mysqli_connect('localhost','root','','pttkhdt'); if (mysqli_connect_errno()) echo "Failed to connect to MySQL: " . mysqli_connect_error(); else { $a = $_SESSION['a'];
<?php
session_start();
$con=mysqli_connect('localhost','root','','pttkhdt');
if (mysqli_connect_errno()) echo "Failed to connect to MySQL: " . mysqli_connect_error();
else {
$a = $_SESSION['a'];
$knad ='SELECT * FROM admintb WHERE adID=' .$a;
$naad = mysqli_query($con,$knad);
$arad = array();
while($rowad=mysqli_fetch_assoc($naad)) $arad[] = $rowad;
}
?>
但是如果我把线路修好
$naad=mysqli_query($con,$knad);
进入:
页面将“消失”并变为空白。使用mysql\u error()检查错误,并在此处更新错误
$naad=mysqli_query($con,$knad) or die(mysql_error());
或者尝试下面的代码
$link=mysql_connect($host,$user,$pwd);
$db = mysql_select_db($dbname,$link);
if(!$db) die (mysql_error());
从外观上看,您正在使用while循环将mysqli_fetch_assoc加载到一个数组中。但是,
mysqli\u fetch\u assoc()
已经在加载数组
你为什么不试试这个:
<?php
session_start();
$con=mysqli_connect('localhost','root','','pttkhdt');
if ( mysqli_errno($con) ) {
//Exit stops the rest of the script
exit( "Failed to connect to MySQL: " . mysqli_error($con) );
} else {
$a = $_SESSION['a'];
$knad = "SELECT * FROM admintb WHERE adID='" . $a . "'";
$arad = mysqli_fetch_assoc( mysqli_query($con,$knad) );
if (!$arad) { exit( mysqli_error($con) ); }
}
?>
只要确保$a
有它的值即可。为什么不使用预先准备好的语句呢,mysqli无论如何都支持这个总和,你能更具体一点吗?我已经在admintb中添加了一行,或者你说的“确保$a
有它的值”是指其他内容吗?这已经很简单了。只要确保$a
有一个值。一旦查询为空,它就会把查询搞砸。要检查,只需使用var\u dump($a)
很抱歉,我的wampserver不允许我使用mysql,我必须改用mysqli,否则它会显示“不推荐的错误”
$link=mysql_connect($host,$user,$pwd);
$db = mysql_select_db($dbname,$link);
if(!$db) die (mysql_error());
<?php
session_start();
$con=mysqli_connect('localhost','root','','pttkhdt');
if ( mysqli_errno($con) ) {
//Exit stops the rest of the script
exit( "Failed to connect to MySQL: " . mysqli_error($con) );
} else {
$a = $_SESSION['a'];
$knad = "SELECT * FROM admintb WHERE adID='" . $a . "'";
$arad = mysqli_fetch_assoc( mysqli_query($con,$knad) );
if (!$arad) { exit( mysqli_error($con) ); }
}
?>