php mysqli_multi_query沒有獲取數據

Question

我嘗試從db進行自動完成搜索的結果。 我使用此代碼獲取名稱，在db中我有一行包含四列可以包含搜索關鍵字。

function search_results($conn,$str){

$sql = "SELECT g_custom_1 as str FROM gallery WHERE  g_custom_1 LIKE '%{$str}%'";
$sql .= "SELECT g_custom_2 as str FROM gallery WHERE  g_custom_2 LIKE '%{$str}%'";
$sql .= "SELECT g_custom_3 as str FROM gallery WHERE  g_custom_3 LIKE '%{$str}%'";
$sql .= "SELECT g_custom_4 as str FROM gallery WHERE  g_custom_4 LIKE '%{$str}%'";


//$sub_data=array();
if (mysqli_multi_query($conn,$sql))
{
    $data=array();
    do
    {
        // Store first result set
        if ($result=mysqli_store_result($conn)) {
        // Fetch one and one row
        while ($row=mysqli_fetch_row($result))
        {
            $data[]=$row['str'];
        }
        // Free result set
        mysqli_free_result($result);
        }
    }
    while (mysqli_next_result($conn));
}
else{
    echo "error";
}
$data_unique=array_unique($data);
return $data_unique;
   }

Answer 1

您需要使用分號分隔每個查詢。 這是您當前的字符串：

$sql="SELECT g_custom_1 as str FROM gallery WHERE  g_custom_1 LIKE '%{$str}%'SELECT g_custom_2 as str FROM gallery WHERE  g_custom_2 LIKE '%{$str}%'SELECT g_custom_3 as str FROM gallery WHERE  g_custom_3 LIKE '%{$str}%'SELECT g_custom_4 as str FROM gallery WHERE  g_custom_4 LIKE '%{$str}%'";

注意后續查詢中的SELECT如何被粉碎到上一個查詢的后端？

我可能會建議：

$sql[]="SELECT g_custom_1 as str FROM gallery WHERE  g_custom_1 LIKE '%{$str}%'";
$sql[]="SELECT g_custom_2 as str FROM gallery WHERE  g_custom_2 LIKE '%{$str}%'";
$sql[]="SELECT g_custom_3 as str FROM gallery WHERE  g_custom_3 LIKE '%{$str}%'";
$sql[]="SELECT g_custom_4 as str FROM gallery WHERE  g_custom_4 LIKE '%{$str}%'";
if(mysqli_multi_query($conn,implode(';',$sql))){

您可能還會發現此答案提供信息： 嚴格標准：mysqli_next_result（）錯誤與mysqli_multi_query