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($
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(
这段代码测试在释放结果之前是否有结果:
$result = mysql_query($query) or die("$query\n\n" . mysql_error());
$id = mysql_insert_id();
if ($result) { mysql_free_result($result); }
但它给出了这样的警告:
mysql_free_result() expects parameter 1 to be resource, boolean given in /foo/bar/etc.php
我做错了什么?
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($
这个问题可能很简单。
$q = "SELECT id FROM users
WHERE name LIKE '%$u%' OR active LIKE '%$u%'";
echo $q;
$u = mysql_query($q) or die("There is a server error : " . mysql_error());
$u = mysql_fetch_array($u);
print_r($u);
但它返回给我
[id] => 6
当我在Mysql Workb
这快把我逼疯了,为什么结果没有显示?
function runSQL($rsql) {
$connect = mysql_connect('localhost','xxx','xxx') or die ("Error: could not connect to database");
$db = mysql_select_db('xxx');
$result = mysql_query($rsql) or die ("Error in query: $query. " .