Question

我正在尝试减少此页面的连接数。我读到的一切都说只有一个连接就足够了但是如果我删除了额外的连接，页面就不会连接到我的服务器并提供我正在寻找的结果。也是我使用两次的查询之一但如果我第二次调用原始查询它也不起作用。

我做错了什么？

<?php
$dbaddress="localhost";
$dbuser="testuser";
$dbpass="testpass";
$dbname="testdb";
$dbtable="elo";
$query="SELECT Sequnce, mcacctname FROM  `elo`;";
$con=mysqli_connect($dbaddress,$dbuser,$dbpass,$dbname);
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }
mysql_connect($dbaddress, $dbuser, $dbpass);
mysql_select_db($dbname);
$sql = "SELECT Sequence, mcacctname FROM  `elo`;";
$result = mysql_query($sql);
mysql_connect($dbaddress, $dbuser, $dbpass);
mysql_select_db($dbname);
$sql2 = "SELECT Sequence, mcacctname FROM  `elo`;";
$result2 = mysql_query($sql2);
$sqlstart = "SELECT mcacctname, elo  FROM  `elo`;";
$q = mysql_query($sqlstart);
?>
<form name="player1" method="post" action="predictions.php">
<label for="Select Player 1">Select Player 1:
<?php
echo "<select name='elouser1'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['Sequence'] . "'>" . $row['mcacctname'] . "</option>";
}
echo "</select>";
?>
</label>
<input type="submit" value="Player 1 Wins">
</form>
<br>
<form name="player2" method="post" action="predictions.php">
<label for="Select Player 2">Select Player 2:
<?php
echo "<select name='elouser2'>";
while ($row2 = mysql_fetch_array($result2)) {
    echo "<option value='" . $row2['Sequence'] . "'>" . $row2['mcacctname'] . "</option>";
}
echo "</select>";
?>
</label>
<input type="submit" value="Player 2 Wins">
</form>
<table>
<tr>
<?
echo '<div class="container">';
while($res = mysql_fetch_array($q)){
echo '<tr><td><div class="item">'.  $res['mcacctname'] . '</td><td>' . $res    ['elo'] . '</div></td></tr>';
}
echo '</div>';
mysqli_close($con);
?>
</tr>
</table>

Answer 1

在页面中打开连接后，您无需为每个查询重新打开它。您可以简单地删除创建连接的所有语句并选择数据库，第一个除外。

注意您正在使用mysqli打开连接，但mysql要查询它。这两组函数不可互换：使用mysqli因为不推荐使用'mysql'，将来会删除支持。

试试这个：

$dbaddress="localhost";
$dbuser="testuser";
$dbpass="testpass";
$dbname="testdb";
$dbtable="elo";
$query="SELECT Sequence, mcacctname FROM  `elo`;";
$con=mysqli_connect($dbaddress,$dbuser,$dbpass,$dbname);
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }
$sql = "SELECT Sequence, mcacctname FROM  `elo`;";
$result = mysqli_query($con,$sql) or die(mysqli_error($con));
$sql2 = "SELECT Sequence, mcacctname FROM  `elo`;";
$result2 = mysqli_query($con,$sql2) or die(mysqli_error($con));
$sqlstart = "SELECT mcacctname, elo  FROM  `elo`;";
$q = mysqli_query($con, $sqlstart) or die(mysqli_error($con));
?>
<form name="player1" method="post" action="predictions.php">
<label for="Select Player 1">Select Player 1:
<?php
echo "<select name='elouser1'>";
while ($row = mysqli_fetch_array($result)) {
echo "<option value='" . $row['Sequence'] . "'>" . $row['mcacctname'] . "</option>";
}
echo "</select>";
?>
</label>
<input type="submit" value="Player 1 Wins">
</form>
<br>
<form name="player2" method="post" action="predictions.php">
<label for="Select Player 2">Select Player 2:
<?php
echo "<select name='elouser2'>";
while ($row2 = mysqli_fetch_array($result2)) {
    echo "<option value='" . $row2['Sequence'] . "'>" . $row2['mcacctname'] . "</option>";
}
echo "</select>";
?>
</label>
<input type="submit" value="Player 2 Wins">
</form>
<table>
<tr>
<?
echo '<div class="container">';
while($res = mysqli_fetch_array($q)){
echo '<tr><td><div class="item">'.  $res['mcacctname'] . '</td><td>' . $res    ['elo'] . '</div></td></tr>';
}
echo '</div>';
mysqli_close($con);
?>
</tr>
</table>

如何减少php连接到mysql服务器的变量和连接数？

1 个答案: