我得到：命令不同步；你现在不能运行这个命令？

Question

我想执行两个查询，但收到“命令不同步；您现在不能运行此命令” 我已经查看了同一问题的先前问题，但无法找出解决方案。我知道这是因为我运行宁两个查询。

谁能告诉我我做错了什么？

查询一：

if($_SERVER["REQUEST_METHOD"] == "POST"){

$sql = "INSERT INTO offer (uniqid, status, l_comp)
VALUES ('test1', 'AFVENTER TILBUD', 'test1');";

$sql .= "INSERT INTO offer (uniqid, status, l_comp)
VALUES ('test2', 'AFVENTER TILBUD', 'test2');";

$sql .= "INSERT INTO offer (uniqid, status, l_comp)
VALUES ('test3', 'AFVENTER TILBUD', 'test3')";

if ($link->multi_query($sql) === TRUE) {

查询二：

$sql = "SELECT * FROM offer_requests WHERE username = '" . 
($_SESSION["username"]) . "'";
                        if($result = mysqli_query($link, 
$sql)){
                            if(mysqli_num_rows($result) > 0){

                                            echo "<th></th>";
                                            echo "<th></th>";
                                            echo "<th></th>";
                                        echo "</tr>";
                                    while($row = 
mysqli_fetch_array($result)){
                                        echo "<tr>";
                                            echo "<th 
class='hidden'>" . $row['uniqid'] . "</th>";
                                            echo "<th>" . 
$row['car'] . "</th>";
                                            echo "<th>" . 
$row['l_comp_one'] . "</th>";
                                            echo "<th>" . 
$row['l_comp_two'] . "</th>";
                                            echo "<th>" . 
$row['l_comp_three'] . "</th>";
                                            echo "<th>";
                                                echo "<a 
href='read_request.php?uniqid=". $row['uniqid'] ."' title='Se 
tilbudsanmodning' data-toggle='tooltip'><span class='glyphicon 
glyphicon-eye-open'></span></a>";
                                            echo "</th>";
                                        echo "</tr>";
                                    }
                                // Free result set
                                mysqli_free_result($result);
                            } else{
                                echo "</br></br><p><b><i>Ingen 
informationer fundet.</i></b></p>";
                            }
                        } else{
                            echo "ERROR: Was not able to 
execute $sql. " . mysqli_error($link);
                        }
                                             echo "</tbody>";                            
                                echo "</table>";

Answer 1

如果我正确理解了另一个答案，这就是该怎么做：

if ($link->multi_query($sql) === TRUE) {
    // the loop below is the fix. It just cycles through the results, ignoring them
    while(mysqli_more_results($link)) {
        mysqli_next_result($link);
        // or $link->next_result();
    }

    //...

我得到：命令不同步；你现在不能运行这个命令？

I get: Commands out of sync; you can't run this command now?

php

mysqli

mysqli-multi-query

我得到：命令不同步；你现在不能 运行 这个命令？

I get: Commands out of sync; you can't run this command now?

php

mysqli

mysqli-multi-query

我得到：命令不同步；你现在不能运行这个命令？