当我试图用php把一个变量放到它的位置时,计数器开始增加小数

x8goxv8g  于 2023-01-19  发布在  PHP
关注(0)|答案(1)|浏览(97)

超文本标记语言

<div class="counter" data-target="
    <?php
    $sql = "select Books_Not_Returned from booksnotreturned ";
    $result = mysqli_query($con, $sql);
    $row = mysqli_fetch_array($result);
    $val = $row['Books_Not_Returned'];
    $Int = (int)$val;
    echo  $Int;
    ?>
    ">0</div>

JS系统

// COUNTER
  const counters = document.querySelectorAll('.counter');
  const speed = 500; // The lower the slower

  counters.forEach(counter => {
  const updateCount = () => {
  const target = +counter.getAttribute('data-target');
  const count = +counter.innerText;

  // Lower inc to slow and higher to slow
  const inc = target / speed;

  // console.log(inc);
  // console.log(count);

  // Check if target is reached
  if (count < target) {
  // Add inc to count and output in counter
  counter.innerText = count + inc;
  // Call function every ms
  setTimeout(updateCount, 1);
  } else {
  counter.innerText = target;
  }
  };

  updateCount();
  });

我试图使计数器停止在数据库中存储的值,当我试图将目标值分配给该变量时,它开始为1.00000000,然后为1.10000,依此类推,直到该变量的实际值

jvlzgdj9

jvlzgdj91#

实际上,在js代码中,我分配const inct = target/speed,因此inc被分配了一个双精度值,以增加延迟,它必须将inc减少到小数,我通过类型转换target/speed修复了它,即将int分配给inc

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