Question

嗨我知道这可能非常简单，经过几个小时的谷歌搜索我已经空手而归，我正在尝试将LIST TABLES查询存储到一个数组中，我可以检查另一个数组中的值是否在生成的LIST表中查询。

$SQL = new DB;

$tables = array(
"Product_Cache",
"Product_Cat",
"Product_Details",
"Product_Images",
"Site_Content",
"UserDetails",
"User_Products",
"User_Type",
"Users"
     );

$result = $SQL->doQuery("SHOW TABLES IN sellmygadgets");
$row = $result->fetch_assoc();

foreach ($tables as $table){
echo "Table " . $table . " Contains this many rows : " . $SQL->numRows(select_all($table)) . "<br>";
    //if(in_array($table, $row, TRUE) {

    //}
}

感谢Andy

，非常感谢帮助

Answer 1

您需要在循环中调用fetch_row()以获取结果的所有行。

$result = $SQL->doQuery("SHOW TABLES IN sellmygadgets");
while ($row = $result->fetch_row()) {
    $table = $row[0];
    if (in_array($table, $tables)) {
        ...
    }
}

Answer 2

    $result = $SQL->doQuery("SHOW TABLES IN sellmygadgets");
while ($row = $result->fetch_row()) {
    $rows[$i] = $row[0];
    $i++;
}

print_r ($rows);

foreach ($tables as $table){    
echo "<br> Table " . $table . " Contains this many rows : " . $SQL->numRows(select_all($table)) . "<br>";

    if (in_array($table, $rows, TRUE)) {
        echo $table . " deos exist <br>";
    } else {
        echo $table . " does not exist <br>";
    }

}

列表表in_array

2 个答案: