![](/img/trans.png)
[英]Console error saying variable not defined, even though defined in same function
[英]Variable is not defined even though it is?
嗨,我試圖在按下按鈕時調用php函數,但標題中始終出現錯誤。
我這樣調用函數:
echo("<th><input type='button' name = 'Attack_Btn' onclick = 'FightPlayer(".$row['username'].")' value ='Attack'></th>");
只是說它是從$ row ['user ...中獲取的用戶名。
index.php:1未捕獲的ReferenceError:未定義Casualjames
這是它接下來調用的代碼
function FightPlayer(enemyName){
var xhttpe;
if (window.XMLHttpRequest) {
xhttpe = new XMLHttpRequest();
} else {
xhttpe = new ActiveXObject("Microsoft.XMLHTTP");
}
xhttpe.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
BattlePlayers();
}
};
xhttpe.open("GET", "FightPlayer.php?enemyname="+enemyName, true);
xhttpe.send();
}
然后它調用我的php腳本,傳入變量enemyname
以供使用
<?php
session_start();
include 'Training.php';
$link = mysqli_connect("","","","");
if (isset($_SESSION['username'])) {
$enemyname = $_REQUEST["enemyname"];
echo $enemyname;
$energyRemove = 1;
$ExperienceGain = 1;
$sql = "SELECT * FROM userstats WHERE username = '$enemyname'";
$result = mysqli_query($link,$sql);
$row = mysqli_fetch_assoc($result);
$Defence = $row["Defence"];
$winChance = CalculateWinChance($link,$Defence);
$sql = "SELECT Energy FROM userstats WHERE username = '".$_SESSION['username']."'";
$result = mysqli_query($link,$sql);
$row = mysqli_fetch_assoc($result);
$rand = rand ( 1 , 100 );
if($row["Energy"] < 1 ){
echo "<script type='text/javascript'>alert('Not enough energy to fight. please restore in character page');</script>";
}else{
if($rand < $winChance){
$_SESSION['Battlemessage'] = "you won against ".$enemyname;
$sql = "UPDATE userstats SET `Energy` = `Energy` - '$energyRemove' WHERE username = '".$_SESSION['username']."'";
mysqli_query($link,$sql);
$sql = "UPDATE userstats SET `Experience` = `Experience` + '$ExperienceGain' WHERE username = '".$_SESSION['username']."'";
mysqli_query($link,$sql);
$sql = "UPDATE userstats SET `Satoshi` = `Satoshi` + 2 WHERE username = '".$_SESSION['username']."'";
mysqli_query($link,$sql);
}else{
$_SESSION['Battlemessage'] = "you lost against ".$enemyname;
$sql = "UPDATE userstats SET `Energy` = `Energy` - '$energyRemove' WHERE username = '".$_SESSION['username']."'";
mysqli_query($link,$sql);
$sql = "UPDATE userstats SET `Satoshi` = `Satoshi` + 1 WHERE username = '".$enemyname."'";
mysqli_query($link,$sql);
}
echo "";
}
calculateLevel($link);
}
?>
我不確定錯誤實際上是在哪里發生的,我已經通過在線代碼檢查器放置了腳本,並且一切都恢復了正常。 我在哪里錯了?
您傳遞給javascript函數的字符串需要用引號引起來,否則它認為這是一個變量:
echo("<th><input type='button' name = 'Attack_Btn' onclick = 'FightPlayer(\"".$row['username']."\")' value ='Attack'></th>");
您的錯誤最有可能與onclick ...有關,您需要在此處的函數參數中轉義引號:
echo("<th><input type='button' name = 'Attack_Btn' onclick = 'FightPlayer(\"".$row['username']."\")' value ='Attack'></th>");
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.