我想从查询中加载联系人,虽然我已经找到了将结果拆分为如下所示的两列的方法,以及它们是一种可点击的button.Is,但它可以显示查询结果,如下面所描述的使用HTML还是查询本身?:
Contacts
------------------------------
Albert Smith | Ben Marshall
Benjamin Jones | Chris Jones
我已经尝试了下面的代码
<?
$rowcount = mysql_num_rows($records);
echo "<div id='column1'>";
$i =
我的db类中有这个方法。
public function query($queryString)
{
if (!$this->_connected) $this->_connectToDb(); //connect to database
$results = mysql_query($queryString, $this->_dbLink) or trigger_error(mysql_error());
return mysql_num_rows($results) > 0 ? mysql_fetch_assoc($
$sql=Select A.a,B.b,b,C.c
from A,B,C
group by
A.a, B.b,C.c
order by A.oder_nr,A.a
$row_Sql= mysql_query($sql,$db) or die (mysql_error());
$record_set= mysql_fetch_assoc($row_Sql);
$totalRows_$row_sql=mysql_num_rows($row_Sql);
$A=''
$B=''
while ( $record_set = mysql_fetch_assoc($
如何将sql结果从一个变量复制到其他没有数组的,并在PHP中预先执行。
$query="Select * from table";
$res1=mysql_query($query);
$res2=$res1; // I need somethink like this, simply, like in other languages
while($row=mysql_fetch_array($res1)) {
echo $row['Id'];
}
echo '<br>';
while($row=mysql_fetch_array($
正在尝试转换为MySQLi。这几天我得到了很大的帮助,谢谢。我的下一个问题是如何在使用LEFT JOIN时进行转换。
运行MYSQL代码
$q1 = 'SELECT * FROM earth LEFT JOIN cities ON (cities.countryid4city = earth.countryid) GROUP BY countryid ORDER by countryid';
$q2 = 'SELECT * FROM cities ORDER BY cityname';
$mytable1 = mysql_query($q1);
$mytable
好的,我试图从mysql表中获取键值的数组,但mysql_fetch_array()似乎不能正常工作。代码:
$x="select id from topics"
$set=mysql_query($x);
echo mysql_num_rows($set);
print_r(mysql_fetch_array($set));
$ids=mysql_fetch_array(
我对mysql_fetch_array()和while循环有问题。我有一个疑问:
$tagsquery = mysql_query("SELECT `url` FROM `tags`, `mapa-tagow`, `statusy` WHERE `tags`.`id` = `mapa-tagow`.`tag-id` AND `statusy`.`id` = `mapa-tagow`.`article-id` AND `tags`.`tag` ='$tag' ORDER BY `url` ASC ") or die("ERROR: Tags doesn