可能重复:
我得到了以下错误:
Warning: mysql_fetch_assoc() expects parameter 1 to be resource
我的查询似乎很好,下面是我的代码:
function products()
{
$query = "SELECT id, quantity, price FROM dvd WHERE quantity > 0";
if (!$query)
{
echo "no product found";
die(mysql_error());
}
我想显示数组的值。但不是展示:
显示ArrayArrayArray.的1509 1510 1511什么意思?
include("db_PSIS.php");
$sql_dc = "SELECT Child, Datecode
FROM traveller_merging15
WHERE Parent='" . $_REQUEST["txt_traveller_no"] . "'
ORDER BY
我想编写一个函数,它包含两个参数:一个是表名,另一个是键的关联数组。现在我有了一个功能:
//返回一个包含$table中所有信息的assoc数组,用于给定的$keyValues function getAllFrom($table,$keyValues){ $qry="SELECT * FROM ". $table; $i=sizeof($keyValues); foreach($keyValues as $key=>$val) { if($i==1) $qry.=" WHERE ".$key."=
我使用以下代码遍历数据库中的几个链接,以检查每个链接的头状态
$con = mysql_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("tracker", $con);
$result=mysql_query
("
SELECT id,ziel_url FROM wm_mapping WHERE z
查看第一个条件//if ($_REQUEST"string"<>'')。<>这个符号是什么意思?这是我第一次看到这样的场景。有人能解释一下这是什么吗..提前感谢
<?php
if ($_REQUEST["string"]<>'') {
$search_string = " AND (full_name LIKE '%".mysql_real_escape_string($_REQUEST["string"])."%' OR e
我有3张桌子:
站点: ID,名称
Assoc: site_id,assoc_type,assoc_id
选项: ID、名称
我想在一个查询中执行以下查询
SELECT * FROM sites
FOREACH site SELECT * FROM assoc WHERE assoc_type='option' AND site_id = site.ID
FOREACH assoc SELECT name FROM options WHERE ID = assoc.ID
FOREACH SITE
echo name, array(option 1, o
错误:
Assoc<float, std::basic_string<char> >&类型引用从const Assoc<float, std::basic_string<char> >类型表达式中的无效初始化
对于此代码
Assoc<keyType,valueType>& found = internalStorage.get(find(key));//returns the value of some key
对不起,我知道这没什么意思,但我很困惑。
有什么问题吗?
在试图从我的MySQL数据库中获取一行(不成功)时,我做了以下操作:
$qry = $this->pdo->prepare("SELECT `post`,`user`,`id` FROM `posts` WHERE `id`='?' ORDER BY `id` DESC");
$qry->execute(array($this->id));
$qry->setFetchMode(PDO::FETCH_ASSOC);
$this->row = $qry->fetch() or die(print_r($qry->e
关于以下代码,我有几个问题:
// Create a new MySQL connection, inserting a host, username, passord, and database you connecting to
$conn = new mysqli('xxx', 'xxx', 'xxx',);
if(!$conn) {
die('Connection Failed: ' . $conn->error());
}
// Create and exe
我得到了这个PHP错误,它是什么意思?
Notice: Undefined offset: 0 in
C:\xampp\htdocs\mywebsite\reddit_vote_tut\src\votes.php on line 41
从下面的代码:
<?php
include("config.php");
function getAllVotes($id)
{
$votes = array();
$q = "SELECT * FROM entries WHERE id = $id";
$r = mysql_quer
我不明白为什么我会犯这个错误:
警告: mysql_fetch_assoc():提供的参数不是有效的MySQL结果资源
我的代码是:
$alreadyMember = mysql_fetch_assoc(mysql_query("SELECT id FROM members WHERE emailAddress='{$_SESSION['register']['email']}'"));
我在这里也得到了错误:
$alreadyRegistered = mysql_fetch_assoc(mysql_query("
CREATE PROCEDURE GetHandTypes()
BEGIN
SELECT * FROM Codemst WHERE code LIKE "HT%" AND dep > 0;
END$$
$result = mysql("CALL GetHandTypes()");
$row = mysql_fetch_assoc($result);
mysql_free_result($result);
它工作得很好。但是当我尝试两次时,它什么也没有返回。
$result = mysql_query("CALL GetHandTypes(
我正在尝试限制数组mysql_fetch_assoc($query),但不确定该如何操作。
$query = mysql_query('SELECT * FROM table ORDER BY id DESC');
while($output = mysql_fetch_assoc($query))
{
//do something
}
你是不是加了个计数器什么的?如何添加此计数器?我对mysql_query和mysql_fetch_assoc真的很困惑。请帮帮我!
因此,我尝试从users表中选择用户名,根据他们是否在组中。
使用的变量:
$member_gang = mysql_fetch_assoc(mysql_query("SELECT level,gangid,rating FROM user_stats WHERE id = '" . $_SESSION['user']['id'] . "'"));
$gang_info = mysql_fetch_assoc(mysql_query("SELECT * FROM gangs WHERE id = '
我正在尝试获取数据库中的所有记录,并将它们逐行显示在表中。它只会一次又一次地显示同一行。如何使表的每一行在数据库中显示下一个结果?
<?php
// Formulate query
$logo = "SELECT logo from stores";
$cat = "SELECT cat from stores";
$commission = "SELECT commission from stores";
$link = "SELECT link from stores";
$name = "SELECT name
这是我的数据库配置,它的扩展是Mysql。我想把它的扩展更改为Mysqli。我想要改变这一点,因为MySQL扩展在最新的php类CMySQL {中不再可用。
// variables
var $sDbName;
var $sDbUser;
var $sDbPass;
var $vLink;
// constructor
function CMySQL() {
$this->sDbName = 'YOUR_DB_NAME';
$this->sDbUser = 'DB_USER_NAME';
$this->sDbPass
<?php
class CMySQL {
// variables
var $sDbName;
var $sDbUser;
var $sDbPass;
var $vLink;
// constructor
public function CMySQL(){
$this->engine = 'mysql';
如何在一个查询中完成以下查询,并以下面的方式获得结果?
// Begining of January
$ob = mysql_query(" SELECT SUM(salary_amount) AS total FROM teacherexpense WHERE month(disburse_date)='01' AND year(disburse_date)='$year' ");
$nt = mysql_fetch_assoc($ob);
$salaryamount= $nt['total'];
$ob = mysq
我找到了这一小段php代码来满足我的需求。
<?php
$allDocs = mysql_query("SELECT id, parent FROM modx_site_content ORDER BY id DESC");
while($doc = mysql_fetch_assoc($allDocs)) {
$parent = mysql_query("SELECT id FROM modx_site_content WHERE parent = " . $doc['id'] . " AND id = " . $doc[
我一直在阅读关于mysql_fetch_*方法的文章。这是我从PHP.org网站上学到的。
mysql_fetch_array — Fetch a result row as an associative array, a numeric array, or both
mysql_fetch_assoc — Fetch a result row as an associative array
mysql_fetch_object — Fetch a result row as an object
mysql_fetch_row — Get a result row as an enumerat
以下代码是使用google api访问的一部分。在connectDb()之后,有一行
$q = sprintf("select * from users where google_user_id='%s' limit 1", r($me->id));
...// and more afterwards
$q = sprintf("insert into users (google_user_id, google_email, google_name, google_picture, google_access_token, created, m
$a = mysql_query("SELECT COUNT(*) as `count_1` FROM `table_1` WHERE `this_1` = '1'");
$b = mysql_fetch_assoc($a);
$c = mysql_query("SELECT COUNT(*) as `count_2` FROM `table_1` WHERE `this_2` = '1'");
$d = mysql_fetch_assoc($c);