Php 数据库返回警告:Printf,无法显示信息
好的,我最后一个工作了,谢谢你们 这是它打印出来的警告,我看不出这里有什么错误 代码现在看起来像这样Php 数据库返回警告:Printf,无法显示信息,php,Php,好的,我最后一个工作了,谢谢你们 这是它打印出来的警告,我看不出这里有什么错误 代码现在看起来像这样 <html> <title>Search Records</title> <head> <body> <form name="" id="" method="post" action="search.php"/> <p> Enter Student name : <input type="t
<html>
<title>Search Records</title>
<head>
<body>
<form name="" id="" method="post" action="search.php"/>
<p> Enter Student name : <input type="text" name="fullname" id="fullname"/>
<input type="submit" name="senda" value="Search Data" />
</form>
<?php
if(isset($_POST['senda'])){
include 'mysqlconn.php';
$con = new mysqli($host, $dbuser, $pass, $db) or die('Cannot Connect');
$name = mysqli_real_escape_string($con,$_POST['fullname']);
$sql = "SELECT * FROM scores WHERE fullname = '$name'";
$result = mysqli_query($con,$sql) or die("Error: ".mysqli_error($con));
while($row = mysqli_fetch_array($result,MYSQLI_ASSOC))
{
printf("%s (%s)\n", $row['Fullname'] ." ". $row['studentNo'] ." ". $row['SubjectName'] ." ". $row['GPA'] ." ". $row['CGPA'] ." ". $row['SCORE']);
mysqli_free_result($result);
}
mysqli_close($con);
}
?>
</body>
</head>
</html>
( ! ) Warning: printf(): Too few arguments in C:\wamp\www\schoolmanagementsystem\search.php on line 24
Call Stack
# Time Memory Function Location
1 0.0051 248384 {main}( ) ..\search.php:0
2 0.1723 265608 printf ( ) ..\search.php:24
( ! ) Warning: mysqli_fetch_array(): Couldn't fetch mysqli_result in C:\wamp\www\schoolmanagementsystem\search.php on line 21
Call Stack
# Time Memory Function Location
1 0.0051 248384 {main}( ) ..\search.php:0
2 0.1901 261928 mysqli_fetch_array ( ) ..\search.php:21
我似乎看不出有什么不对劲,我到底做错了什么