doudou8081 2016-06-08 06:33
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为什么一个php标签内的两个查询无法正常工作

I have tried out some code were two mysql statement need to be executed, I don't have any syntax error though, but second mysql query is not working and not giving out any results. Please help me out am pretty new to this field if any mistake pardon me, thank you.

.php

<?php
$con = mysqli_connect("localhost", "*****", "*****", "******");
$query = ("SELECT * FROM profile");
$result = mysqli_query($con, $query);
while ($row = $result->fetch_assoc()) {
    $query1 = ("SELECT vault_no FROM  grp_tbl");
    $result1 = mysqli_query($con, $query1);
    while ($row1 = $result1->fetch_assoc()) {
        if ($row1['vault_no'] !== $row['vault_no']) {
            echo
            '<div class = "chat-user-name"><span class="pull-right label label-primary"><input name="ppl" type="radio" value= "' . $row["via"] . '" "></span><div align="center"><input type="hidden" name="category" value="macro">
                                                </div>
                                                ' . $row['via'] . ' </div>';
        }
    }
}
?>
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3条回答 默认 最新

  • douzhouqin6223 2016-06-08 07:07
    关注

    Use like this, it may work

    <?php
    $con = mysqli_connect("localhost", "*****", "*****", "******");
    $query = ("SELECT profile.* from profile left join grp_tblon profile.vault_no!=grp_tbl.vault_no where profile.vault_no!=grp_tbl.vault_no");
    $result = mysqli_query($con, $query);
    while ($row = $result->fetch_assoc()) {
                echo
                '<div class = "chat-user-name"><span class="pull-right label label-primary"><input name="ppl" type="radio" value= "' . $row["via"] . '" "></span><div align="center"><input type="hidden" name="category" value="macro">
                                                    </div>
                                                    ' . $row['via'] . ' </div>';
        }
    }
    ?>
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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