我使用火狐也一樣,我有同樣的問題發展我輸入型數字輸入caracters和空格等... 我已經找到了標籤模式的解決方案=「[1- 9]「,但不幸的是它不適合我。在沒有任何結果的情況下進行無謂的搜索之後,我決定開發自己的功能。 反正即時通訊在這個例子採用了棱角分明2,它幾乎類似的JavaScript,這樣你就可以在任何情況下使用此代碼: 這裏是HTML:
<input class="form-control form-control-sm" id="qte" type="number" min="1" max="30" step="1" [(ngModel)]="numberVoucher"
(keypress)="FilterInput($event)" />
下面是函數FilterInput:
FilterInput(event: any) {
let numberEntered = false;
if ((event.which >= 48 && event.which <= 57) || (event.which >= 37 && event.which <= 40)) { //input number entered or one of the 4 directtion up, down, left and right
//console.log('input number entered :' + event.which + ' ' + event.keyCode + ' ' + event.charCode);
numberEntered = true;
}
else {
//input command entered of delete, backspace or one of the 4 directtion up, down, left and right
if ((event.keyCode >= 37 && event.keyCode <= 40) || event.keyCode == 46 || event.which == 8) {
//console.log('input command entered :' + event.which + ' ' + event.keyCode + ' ' + event.charCode);
}
else {
//console.log('input not number entered :' + event.which + ' ' + event.keyCode + ' ' + event.charCode);
event.preventDefault();
}
}
// input is not impty
if (this.validForm) {
// a number was typed
if (numberEntered) {
let newNumber = parseInt(this.numberVoucher + '' + String.fromCharCode(event.which));
console.log('new number : ' + newNumber);
// checking the condition of max value
if ((newNumber <= 30 && newNumber >= 1) || Number.isNaN(newNumber)) {
console.log('valid number : ' + newNumber);
}
else {
console.log('max value will not be valid');
event.preventDefault();
}
}
// command of delete or backspace was types
if (event.keyCode == 46 || event.which == 8) {
if (this.numberVoucher >= 1 && this.numberVoucher <= 9) {
console.log('min value will not be valid');
this.numberVoucher = 1;
//event.preventDefault();
this.validForm = true;
}
}
}
// input is empty
else {
console.log('this.validForm = true');
this.validForm = false;
}
};
在這個功能我不得不讓數字,方向,刪除按鍵進入。
對於沒有JavaScript的Firefox,這是不可能的。如果你錯了,並且鍵入任何非數字符號 - 字段,那麼該字段仍然有效,在這種情況下它不被驗證。 如果你需要例子,如何用JS做到這一點,讓我知道。 –
默認情況下,Mac OS X中的Firefox 39不會阻止字母字符輸入,並且Firefox 42僅會驗證,但不會禁用字母鍵輸入。 推薦來源:http://caniuse.com/#feat=input-number(已知問題標籤) – Venkat