Question

这是我的A-level comp sci课程的一些课程。我目前正在使用一种非常陌生的语言进行编码，我需要一些方面的帮助，我现在已经苦苦挣扎了一段时间。下面显示了我到目前为止所做的工作。我想要做的是输出表中对应的所有值都有一个特定的运动ID这是我到目前为止所拥有的，但它并没有返回任何值，尽管填充了表。

<?php
$link = mysqli_connect("localhost", "root", "pizza","fixtures");

if ($_POST['SPORT'] == "Football") {
    $sp = '1';
}
if ($_POST['SPORT'] == "Tennis") {
    $sp = '2';
}
if ($_POST['SPORT'] == "Swimming") {
    $sp = '3';
}

$result = mysql_query("SELECT * FROM fixtureDetails WHERE fixtureDetails.sportID = '$sp'");
if(mysqli_num_rows($result) > 0) {
    echo "yes";
}
mysqli_close($link);
?>

Answer 1

您的代码存在一些问题：

不要混合多个图书馆。您正在使用mysql_和mysqli_。
不要使用mysql_*个功能。它们已被弃用。
无需mysqli_close()功能。
你不需要重复这张桌子。
您没有打印查询结果中的任何内容。

您的代码存在问题，您需要更改此行：

$result = mysqli_query($link, "SELECT * FROM `fixtureDetails` WHERE `sportID`='$sp'");

要填充为表格，请使用结果集和循环。

if (mysqli_num_rows($result)) {
  while (false != ($data = mysqli_fetch_assoc($result))) {
    // Do whatever with your data.
    var_dump($data);
  }
} else {
  echo "No records.";
}

最终代码

<?php
  $link = mysqli_connect("localhost", "root", "pizza","fixtures");

  if ($_POST['SPORT'] == "Football") {
    $sp = '1';
  }
  if ($_POST['SPORT'] == "Tennis") {
    $sp = '2';
  }
  if ($_POST['SPORT'] == "Swimming") {
    $sp = '3';
  }

  // Execute the query and save the resultset.
  $result = mysqli_query($link, "SELECT * FROM `fixtureDetails` WHERE `sportID`='$sp'");
  // Check if there are any rows returned.
  if (mysqli_num_rows($result)) {
    // If there are rows returned, save every row to $data.
    while (false != ($data = mysqli_fetch_assoc($result))) {
      // Do whatever with your data.
      var_dump($data);
    }
  } else {
    // If there are no records, display a message.
    echo "No records.";
  }
?>

如果你想让一个函数发送计数的响应，你可以这样：

<?php
  function getCount() {
    $link = mysqli_connect("localhost", "root", "pizza","fixtures");

    if ($_POST['SPORT'] == "Football") {
      $sp = '1';
    }
    if ($_POST['SPORT'] == "Tennis") {
      $sp = '2';
    }
    if ($_POST['SPORT'] == "Swimming") {
      $sp = '3';
    }

    // Execute the query and save the resultset.
    $result = mysqli_query($link, "SELECT * FROM `fixtureDetails` WHERE `sportID`='$sp'");
    // Check if there are any rows returned.
    return mysqli_num_rows($result);
  }
?>

如果您只想要true或false，则可以执行以下操作：

<?php
  function getCount() {
    $link = mysqli_connect("localhost", "root", "pizza","fixtures");

    if ($_POST['SPORT'] == "Football") {
      $sp = '1';
    }
    if ($_POST['SPORT'] == "Tennis") {
      $sp = '2';
    }
    if ($_POST['SPORT'] == "Swimming") {
      $sp = '3';
    }

    // Execute the query and save the resultset.
    $result = mysqli_query($link, "SELECT * FROM `fixtureDetails` WHERE `sportID`='$sp'");
    // Check if there are any rows returned.
    return (mysqli_num_rows($result) > 0);
  }
?>

如何检查数据库中是否存在值，然后返回数据行

1 个答案: