Warning: file_get_contents(/data/phpspider/zhask/data//catemap/1/php/252.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
php mysql查询| html表_Php_Mysql_Html Table - Fatal编程技术网

php mysql查询| html表

php mysql查询| html表,php,mysql,html-table,Php,Mysql,Html Table,伙计们,我现在很困惑,表中填充的数据没有显示第一条记录 html表格: <table class="table table-bordered table-condensed" align="center" bordercolor="#CCCCCC"> <tr bgcolor="#009933"> <td align="center" style="col

伙计们,我现在很困惑,表中填充的数据没有显示第一条记录

html表格:

<table class="table table-bordered table-condensed" align="center" bordercolor="#CCCCCC">
                            <tr bgcolor="#009933">
                            <td align="center" style="color:#FFF;">Name</td>
                            <td align="center" style="color:#FFF;">Course</td>
                            <td align="center" style="color:#FFF;">Grade</td>
                            <td align="center" style="color:#FFF;">Remark</td>
                            </tr>
                            <?php
                            while($result= mysql_fetch_array($query1)){
                                echo "<tr>";
                                echo "<td>".$result['stud_name']."</td>";
                                echo "<td class=\"text-center\">"."</td>";
                                echo "<td>".$result['grade']."</td>";
                                echo "<td>".$result['remark']."</td>";
                            }
                            ?>
</table>
但是当我使用json测试查询时,它会显示所有记录

{
    "test": [{
        "stud_name": "Jeo De Jesus",
        "grade": "58",
        "remark": "Failed"
    }, {
        "stud_name": "Juana Gonzales",
        "grade": "60",
        "remark": "Failed"
    }, {
        "stud_name": "Wendy Lizardo",
        "grade": "81",
        "remark": "Passed"
    }, {
        "stud_name": "Jeffrey Oliveras",
        "grade": "91",
        "remark": "Passed"
    }, {
        "stud_name": "Mc Jester Salinas",
        "grade": "83",
        "remark": "Passed"
    }],
    "success": 1
}

我认为你没有结束你的工作。试试这个。(未测试)


名称
课程
等级
评论

您尚未关闭while循环中的
标记。我不确定它是否能解决您的问题,但它不正确。

正如您所建议的,先生,我关闭了它,但仍然没有任何改变:(省略
结束标记不是错误。它不会导致问题。好的,我解决它^ ^它与此线程有关=>问题可能是您正在使用mysql,它在旧版本中被降级,并从新的php版本中删除,所以请尝试使用mysqli或PDO,为了省略sql注入风险,请使用prepared语句扫描您。)如果您是mysql\u fetch\u数组($query1)结果,则匹配
结束标记不是错误。这不会导致问题。我认为您是在mysql\u fetch\u数组($query1)之前执行sql。请共享您的所有代码,以便我们检查。
{
    "test": [{
        "stud_name": "Jeo De Jesus",
        "grade": "58",
        "remark": "Failed"
    }, {
        "stud_name": "Juana Gonzales",
        "grade": "60",
        "remark": "Failed"
    }, {
        "stud_name": "Wendy Lizardo",
        "grade": "81",
        "remark": "Passed"
    }, {
        "stud_name": "Jeffrey Oliveras",
        "grade": "91",
        "remark": "Passed"
    }, {
        "stud_name": "Mc Jester Salinas",
        "grade": "83",
        "remark": "Passed"
    }],
    "success": 1
}
<table class="table table-bordered table-condensed" align="center" bordercolor="#CCCCCC">
                        <tr bgcolor="#009933">
                        <td align="center" style="color:#FFF;">Name</td>
                        <td align="center" style="color:#FFF;">Course</td>
                        <td align="center" style="color:#FFF;">Grade</td>
                        <td align="center" style="color:#FFF;">Remark</td>
                        </tr>
                        <?php
                        while($result= mysql_fetch_array($query1)){
                            echo "<tr>";
                            echo "<td>".$result['stud_name']."</td>";
                            echo "<td class=\"text-center\">"."</td>";
                            echo "<td>".$result['grade']."</td>";
                            echo "<td>".$result['remark']."</td></tr>";
                        }
                        ?>