【发布时间】:2009-04-01 22:40:59
【问题描述】:
您好,我尝试创建一个函数来生成选择函数。
但是下面的代码
public function select($psTableName, $paFields ="",$paWhere=array())
{
//initial return value
$lbReturn = false;
try
{
$lsQuery = "SELECT * FROM `";
$lsQuery .= $psTableName;
$lsQuery .= "` ";
if (!empty($paWhere)){
$lsQuery .= "WHERE ";
print_r($paWhere);
foreach($paWhere as $lsKey => $lsValue)
{
echo $lsKey."<br>";
$paWhere[] = $lsKey." = '".mysql_real_escape_string($lsValue)."'";
}
$lsQuery .= implode(" AND ", $paWhere);
//$lsQuery = substr($lsQuery,0,(strlen($lsQuery)-5));
}
//echo $lsQuery;
//execute $lsQuery
$this->msLastQuery = $lsQuery;
if(!$this->execute())
{
throw new DBException("Select failed, unable to execute query: ".$lsQuery);
}
else
{
$lbReturn = true;
}
}
catch(DBException $errorMsg)
{
ErrorHandler::handleException($errorMsg);
}
return $lbReturn;
}
生成这条 sql 语句:
SELECT * FROM `persons` WHERE email@gmail.com AND 2d1cf648ca2f0b2499e62ad7386eccc2 AND 1 AND per_email = 'email@gmail.com' AND per_password = '2d1cf648ca2f0b2499e62ad7386eccc2' AND per_active = '1'
我不知道为什么它首先只显示 where 子句之后的值,然后返回并显示 key => 值。
有什么想法吗?
【问题讨论】:
标签: php select associative-array