在代码中,我从数据库收集数据,然后根据数据的值更新它。我可以检索这些值,但无法更新它们。我需要有关更新函数或不同类型函数的语法的帮助。
<?php
$con = mysqli_connect("127.0.0.1","root","","timeclock");
if (mysqli_connect_errno($con))
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}else{
echo "connected";
}
$x=2;
timein($x);
function timein($x){
$con = mysqli_connect("127.0.0.1","root","","timeclock");
if (mysqli_connect_errno($con))
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}else{
echo "connected";
}
$result=mysqli_query($con,"SELECT * FROM emplo`enter code here`yeedatabase WHERE employeeID=$x ORDER BY employeeID" );
while($row = mysqli_fetch_array($result))
{
$inout = $row['inOut'];
}
echo $inout;
if($inout == 1){
echo "hello";
$con = mysqli_connect("127.0.0.1","root","","timeclock");
if (mysqli_connect_errno($con))
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}else{
echo "connected";
}
$result=mysqli_query($con,"SELECT * FROM employeedatabase WHERE employeeID=$x ORDER BY employeeID" );
while($row = mysqli_fetch_array($result))
{
$inout = $row['inOut'];
$row['inOut']=2;
}
echo "</table>";
echo $inout;
$outa=2;
mysqli_query($con,"UPDATE employeedatabase SET inOut='$outa' WHERE employeeID='$x'");
}
}
?>-## Heading ##
mysqli 查询函数不起作用,我的 mysqli 函数需要一个新函数或更好的语法。
最佳答案
“不工作”不是很有帮助,在将来的错误以及放入的内容和发生/不发生的情况下是有帮助的。
但看起来这条线需要改变
SELECT * FROM emplo`enter code here`yeedatabase WHERE employeeID=$x ORDER BY employeeID" );
至
SELECT * FROM employeedatabase WHERE employeeID=$x ORDER BY employeeID" );
关于php - 我正在寻找新的更新功能或建议来修复我现有的功能,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/19369096/