我有一个表,其中包含城市,人,就业状态.我想使用PHP使用此值填充html表,第一个td将包含一个城市,下一个将包含该城市的总人数,而第三个td将包含雇用的总人数.通常,我使用while填充表,但是这一步我不知道如何进行mysqli查询,如果while也可以使用,请检查解决方案,但找不到,在这里呆了几天.
I have a table that contains city, person, employment_status. I want to populate my html table with this values using PHP, the first td will contain a city the next will contain the total number of people in that city and the third td will contain the total number of people employed. Normally I populate my table using while but this one I don't know how to go about the mysqli query and if while will also work, checked for a solution but can't find, been stuck here for days.
<table class="table table-striped table-advance table-hover search-table" > <tbody> <thead> <tr> <th><i class="icon_profile"></i> city</th> <th><i class="icon_pin_alt"></i> Total number of people</th> <th><i class="icon_pin_alt"></i> no of pepople employed</th> </tr> </thead> <?php $sql = "SELECT COUNT(DISTINCT city) FROM user"; $q=$conn->query($sql); while ($row = mysqli_fetch_array($q)) {卡在这里,它是查询的一部分,在db中没有创建向我提供此值的行
am stuck here, the query part of it, there is no row created in db to provide me this values
推荐答案人们在评论中建议.您必须显示到目前为止所做的一切.但是这段代码可能会有所帮助.我正在使用mysqli连接.
As the people suggested in the comments. You'll have to show what have you done so far.But this code may come helpful.I am using using mysqli connection.
<table> <thead> <tr> <th>City</td> <th>Person</td> <th>Employment Status</td> </tr> </thead> <tbody> <?php $SELECT = mysqli_query($conn,"SELECT * FROM `table`"); if($SELECT != false) { while($rows = mysqli_fetch_array($SELECT)) { echo " <tr> <td>".$rows["city"]."</td> <td>".$rows["person"]."</td> <td>".$rows["employment_status"]."</td> </tr> "; } } else { echo " <tr> <td colspan='3'>Something went wrong with the query</td> </tr> "; } ?> </tbody> </table>更多推荐
使用数据库值填充HTML表
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