我srSignals在一个名为的数据库中有一个表,signals如下所示:
animal inserttime id
dog 1970-00-00 00:00:00 3042
cat 1970-00-00 00:00:00 3041
monkey 1970-00-00 00:00:00 3040
hamster 1970-00-00 00:00:00 6
我的所有inserttime价值观都被重置为1970年.幸运的是,我signals使用正确的inserttime值(和相应的id值)备份数据库.调用备份文件signals.sql.gz(signals.sql显然在gz压缩内部).
我的问题是这样的:使用phpMyAdmin,我怎么能update列inserttime在id现有的表=的id备份文件.请注意,我不想完全恢复表,因为在过渡期间已将其他内容添加到数据库中.
我不确定最好的方法 - 恢复phpMyAdmin的功能？一些花哨的更新/加载？
谢谢!
嗯,我是一个考虑PHP/HMTL编程的绝对初学者,我现在已经坚持这些问题了几天.
首先,我创建了包含以下列的mysql表(id,Name,Surname,Telephone_number,Email).然后我成功创建了阅读表格.现在我想阅读并稍后使用以下代码编辑特定行:
41 <?php
42 $con = mysql_connect("localhost","username","password");
43 mysql_select_db("database", $con);
44 $id=$_GET["id"];
45 $result = mysql_query("SELECT * FROM Table where id='$id'");
46 $row = mysql_fetch_array($result);
47 ?>
但我得到的是以下错误:注意:未定义的索引:第44行的C:\ xampp\htdocs\contacts_edit.php中的id
任何帮助将不胜感激!
这是我的html表单的代码:
<!DOCTYPE html>
<head>
<title> Question </title>
<style type = "text/css">
body {
font-family:cursive;
}
a:link {
text-decoration:none;
background-color:#D0D0D0;
color:#000000;
width:100px;
display:block;
text-align:center;
padding:4px;
}
a.visited {
text-decoration:none;
background-color:#D0D0D0;
color:#000000;
width:100px;
display:block;
text-align:center;
padding:4px;
}
a.active {
text-decoration:none;
background-color:#D0D0D0;
color:#000000;
width:100px;
display:block;
text-align:center;
padding:4px;
}
a:hover {
background-color:#686868;
color:#FFFFFF;
}
#title {
text-align:center;
}
</style>
</head>
<body>
<?php
session_start();
?>
<h1 id="title"> Question 1 </h1>
<br/>
<form action="q15.php" method="POST" >
<fieldset>
<legend>Who wrote the music we all recognise from the …下面的代码似乎没有正常工作,当我在WHERE中只放一件东西它工作正常,所以也许我不能在这里做多个项目？
代码片段是
$query_view_users = "SELECT id, type, name, username, email FROM admin_users WHERE id, type, name, username, email LIKE '$search'";
非常感谢你的帮助.:-)
我正在尝试建立一个登录过程,在我的phpmyadmin中连接我的AdminLogin2.php和admin2表,但它表示拒绝访问.我不知道该怎么做才能解决它.希望有人可以帮助我.在这里我的代码,谢谢.
`<?php
$host="localhost";
$username="root";
$password="";
$database="finalproject";
$table="admin2";
$AdminID=$_POST['Field1'];
$Password=$_POST['Field3'];
mysql_connect("$host","$password") or die (mysql_error());
#echo"connected";
mysql_select_db("$database") or die (mysql_error());
#echo"database found";
// To protect MySQL injection (more detail about MySQL injection)
$AdminID = stripslashes($AdminID);
$Password = stripslashes($Password);
$AdminID = mysql_real_escape_string($AdminID);
$Password = mysql_real_escape_string($Password);
$sql="SELECT * FROM $admin2 WHERE username='$AdminID' and password='$Password'";
$result=mysql_query($sql);
// Mysql_num_row is counting table row
$count=mysql_num_rows($result);
// If result matched $myusername and $mypassword, table row must be 1 row
if($count==1){
// Register …如果我删除phpmyadmin上的mySQL数据库会影响我的ftp服务器上传的内容吗？mySQL表和FTP内容是否以某种方式相关？我正在使用wordpress来建立网站.
这是我的查询,当我在phpmyadmin中运行此查询时,它检索到正确的结果但是当我在php中运行它时会出现这种错误"注意:未定义的索引:pcat在C:\ xampp\htdocs\classified\index.php ".我不明白这个错误,因为"pcat"是数据库中的一个列,它在查询中定义,为什么会出现这种错误？请帮我
<?php
$query = mysql_query("SELECT COUNT( pcat ) FROM category WHERE pcat = 'Jobs'") or die(mysql_error());
while($getcount = mysql_fetch_array($query)){
?>
<span class="count">(<?php echo $getcount['pcat']; ?>)</span></th>
<?php } ?>
您好我怎么能将这种文本文件导入MySQL数据库感谢魁北克市提前:)
"8638671",1,1,"SAL",,,"12.8x16.9 P",,"RC","PFLO",,,
"8638671",2,2,"CUI",,,"10.11x12.6 P",,"RC","LINO",,,
"8638671",3,3,"SDB",,,"3.11x7.3 P",,"RC","LINO",,,
"8638671",4,4,"CCP",,,"8.6x10.2 P",,"RC","LINO",,,
"8638671",5,5,"SDL",,,"5.6x4.4 P",,"RC","LINO",,,
"8640420",1,1,"HAL",,,"5.5x3.3 P",,"RC","CERAM",,,
"8640420",2,2,"SAL",,,"11.1x21.8 P",,"RC","PFLO",,,
"8640420",3,3,"SAM",,,"9x11 P",,"RC","PFLO","porte-patio",,
"8640420",4,4,"CUI",,,"8x7.5 P",,"RC","CERAM",,,
"8640420",5,5,"SDB",,,"5x8 P",,"RC","CERAM",,,
"8640420",6,6,"SDL",,,"5x7.5 P",,"RC","LINO",,,
"8640420",7,7,"CCP",,,"11.5x13 P",,"RC","PFLO",,,
"8640420",8,8,"CAC",,,"9x10 P",,"RC","PFLO",,,
"8640420",9,9,"CAC",,,"7.5x9.10 P",,"RC","PFLO",,,
"8672024",1,1,"CUI",,,"10x8 P",,"RC","AU",,,
"8672024",2,2,"SAM",,,"15x10 P",,"RC","AU",,,
我遇到了带有日期条件的select queryI(下面)的问题.查询没有给出正确的结果,我不知道为什么？
请帮我谢谢.在2013-01-01到2013-01-31之间,数据库中的Created_Date字段有值
SELECT * FROM (`cms_product`) WHERE `cms_product`.`Created_Date` >= 2013-01-01
and `cms_product`.`Created_Date` <= 2013-01-31 ;
Created_Date字段位于Datetime中.
我无法使用phpMyAdmin创建一个函数,就像MySQL的界面一样.所以我写道:
DELIMITER //
Create or Replace Function affecter (id_patient IN integer , id_maladie IN VARCHAR2 )
Return VARCHAR2(30) IS msg Varchar2(30);
Begin
Insert into souffrir values (id_patient,id_maladie);
msg:= 'Insertion effectuée';
Return msg;
END//
我收到此错误:
#1064 - 您的SQL语法有错误; 查看与您的MySQL服务器版本对应的手册,以便在'功能影响器(id_patient IN integer,id_maladie IN VARCHAR2)附近使用正确的语法.在第1行返回V'
我该如何解决这个错误？