【问题标题】:in if condition both equal to and not equal to returns true for the same condition在 if 条件中,等于和不等于对于相同的条件返回 true
【发布时间】:2020-03-24 15:49:01
【问题描述】:
<div class="row">
    <div class="container">
        <div class="col s12">
            <ul class="pagination center-align text-blue">

            <?php 

                $get_pagCount = "SELECT COUNT(p_id) as count_pro FROM products";

                $run_pagCount = mysqli_query($con,$get_pagCount);

                $row_pagCount = mysqli_fetch_array($run_pagCount);

                $totalRecords = $row_pagCount["count_pro"];

                $totalPages = ceil($totalRecords / $perPage);               

            ?>  

            <!-- left symbol of pagination -->          

            <?php if ($paginationNo==1): ?>

                <li class="waves-effect disabled"><a href="#!"><i class="material-icons">chevron_left</i></a></li>

            <?php endif; ?>

            <?php if ($paginationNo!==1): ?>

                <li class="waves-effect"><a href="products.php?paginationId=<?php echo($paginationNo-1); ?>"><i class="material-icons">chevron_left</i></a></li>

            <?php endif; ?>


            <!-- pagination numbers dynamics -->

            <?php for ($i=1; $i<=$totalPages; $i++): ?>

                <?php if ($paginationNo == $i): ?>

                    <li class="waves-effect active"><a href="#!"><?php echo($i); ?></a></li>

                <?php endif; ?>

                <?php if ($paginationNo !== $i): ?>

                    <li class="waves-effect"><a href="products.php?paginationId=<?php echo($i); ?>">2</a></li>

                <?php endif; ?>

            <?php endfor; ?>




            <!-- right symbol of pagination -->

            <?php if ($paginationNo == $totalPages): ?>

                <li class="waves-effect disabled"><a href="#!"><i class="material-icons">chevron_right</i></a></li>

            <?php endif; ?>

            <?php if ($paginationNo !== $totalPages): ?>

                <li class="waves-effect"><a href="products.php?paginationId=<?php echo($paginationNo+1); ?>"><i class="material-icons">chevron_right</i></a></li>


            <?php endif; ?>

            </ul>
        </div>
    </div>
</div>

上面代码中所有等于和不等于的if条件都对相同的条件返回true。这怎么可能?如下图所示,为分页的左箭头创建了两个箭头,为 1 个分页创建了两个 li,一个为活动的,另一个为不活动的,如上面代码中的 if 条件所述。如果我使 $paginationNo = 5 等于 $totalPages 则左箭头创建一次,但分页右箭头创建两次 output of the code above

【问题讨论】:

标签: php mysql if-statement


【解决方案1】:

“试试 != 因为我假设 $paginationNo 可能是一个字符串,1 是数字,因此测试在数据类型上失败”

【讨论】:

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