相关疑难解决方法(0)

我正在尝试运行以下内容.

<?php

$db = mysqli_connect("localhost","user","pw") or die("Database error");
mysqli_select_db($db, "database");

$agtid = $_POST['level'];

$sql = sprintf("call agent_hier(%d)", $agtid);

$result = mysqli_query($db, $sql) or exit(mysqli_error($db));

if ($result) {
    echo "<table border='1'>
        <tr><th>id</th>
        <th>name</th>
        <th>parent_id</th>
        <th>parent_name</th>
        <th>level</th>
        <th>email</th></tr>";

    while ($row = mysqli_fetch_assoc($result)) 
    {
        $aid = $row["id"];
        $sql2 = "SELECT * FROM members WHERE MEMNO = '$aid'";
        $result2 = mysqli_query($db,$sql2) or exit(mysqli_error($db));

            while ($newArray = mysqli_fetch_array($result2)) {
                $fname = $newArray['FNAME'];
                $lname = $newArray['LNAME'];
                $mi = $newArray['MI'];  
                $address = $newArray['ADDRESS'];    
                $city = $newArray['CITY'];  
                $state …

我正在尝试运行一个程序,我收到此错误

Commands out of sync; you can't run this command now

这是我得到的原始错误

命令不同步; 你现在不能运行这个命令

SELECT DISTINCT `property_id`, `pin`, `block_id`, `serial_no`, `status`, `ex_sn`, `ex_code`, `property_date_time`, `street_add`, `lab_name` FROM `view_property_user_lab` WHERE status = '6' AND lab_id = '01' AND designation IN( '5','6') LIMIT 10 

可以任何1告诉我为什么我得到这个错误以及如何摆脱它.我正在使用Code igniter,我也试过这个

$query->free_result().

在我的程序中,我使用了这个陈述

   SELECT *
   FROM
  temp_calculated_rates_and_rules;
 -- and then
   TRUNCATE temp_calculated_rates_and_rules;

因为这个东西在PHP循环中被调用,就像这样

  $arrIds = array('5','10');
    foreach ($arrIds as $id)
    {
        $this->_StoredProcedureMapper->setPId($id);

        $p10values = $this->_StoredProcedureMapper->fetch_p10_values();
        if (intval(@$p10values[0]['is_exempted']) != 1)
        {
            $this->generate_p10($p10values);
        }

    }

这是mapper函数

    function fetch_p1_values()
{ …