Question

运动选项的类型是从我拥有各种运动类型的数据库中选择的。 Type_sport包含运动的名称。一切都很好，数据库中的所有体育都显示在选项列表中。但是，当我从数据库中选择选项时，我如何$ _REQUEST用户选择了哪种类型的运动？表单中的代码如下所示：

do
$$
declare
 _i text;
 _r text;
begin
 _i := '${table_1} + ${table_2} + ${table_3}';
 _r := (
   with u as (select * from unnest(string_to_array(_i,'+')) with ordinality t(e,o))
  , c as (select format('TMP_%s AS (SELECT date, quantity FROM %I)'
  , o
  , substring(unnest(string_to_array(e,'+')) from '\{(.+)\}')
  ) from u
)
select string_agg(format,','||chr(10)) from c);
 _r := 'WITH '||chr(10)||_r||chr(10)||$s$SELECT time_bucket('1 day', date) AS date,
    sum(quantity) AS sum FROM
    (
    SELECT date, quantity FROM TMP_1
    UNION ALL
    SELECT date, quantity FROM TMP_2
    UNION ALL
    SELECT date, quantity FROM TMP_3
    ) AS res_1
  GROUP BY 1
  ORDER BY 1;$s$;
 raise info '%',_r;
 execute _r;
end;
$$
;

我尝试过：<p>Choose sport: </p> <form action="page2.php" method="post"> <select name="sports"> <option>Choose sport</option> <?php $db=mysqli_connect("xxxxxxxx","xxxxx","","xxxxx"); $sql="Select* from sports"; $result=mysqli_query($db,$sql); $numberOfRows=mysqli_affected_rows($db); for($i=0;$i<$numberOfRows;$i++){ $row=mysqli_fetch_array($result); echo "<option> ".$row["type_sport"]."</option>"; } ?> </select> </form>但它不起作用。 `

Answer 1

您需要为选项标记提供一个值属性，这是您从请求超级全局it('should load the school results in less than 1s', () => { searchPage.navigateTo(); browser.waitForAngularEnabled(false); searchPage.search('col'); browser.wait(EC.visibilityOf(searchPage.results), 1000); expect(searchPage.resultLinks.count()) .toBeGreaterThan(0); });

获取值的唯一方式

我如何$ _REQUEST数据库做出的选项？

1 个答案: