如何在php回音上引用回音div的javascript函数
我试图在php上回显这段代码,但由于有许多代码比它应该的时间更早结束html,我如何修复这段代码如何在php回音上引用回音div的javascript函数,javascript,php,html,css,Javascript,Php,Html,Css,我试图在php上回显这段代码,但由于有许多代码比它应该的时间更早结束html,我如何修复这段代码 function button($conn){ $sql = "SELECT * FROM table"; $result= mysqli_query($conn, $sql); while($row = $result->fetch_assoc()){ echo "<button class='FASTFONT' onclick='openCity(e
function button($conn){
$sql = "SELECT * FROM table";
$result= mysqli_query($conn, $sql);
while($row = $result->fetch_assoc()){
echo "<button class='FASTFONT' onclick='openCity(event,'".$row['name']."')'>".$row['name']."</button>";
}
}
这是我得到的输出:
尝试使用\而不是常规的“尝试这样做:
function button($conn){
$sql = "SELECT * FROM table";
$result= mysqli_query($conn, $sql);
while($row = $result->fetch_assoc()){
echo '<button class="FASTFONT" onclick="openCity(event,\''.$row['name'].'\')">'.$row['name'].'</button>';
}
}
需要更改echo语句,以便正确转义与开始和结束引用相匹配的引用。这里有一种方法可以做到这一点:
function button($conn){
$sql = "SELECT * FROM table";
$result= mysqli_query($conn, $sql);
while($row = $result->fetch_assoc()){
echo "<button class='FASTFONT' onclick='openCity(event,\"".$row['name']."\")'>".$row['name']."</button>";
}
}
输出应该是这样的:
<button class='FASTFONT' onclick='openCity(event,"Blanda1")'>Blanda1</button>
在echo语句中使用$name=$row['name'],并使用$name。
<button class='FASTFONT' onclick='openCity(event,"Blanda1")'>Blanda1</button>